Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.

There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.

We have to note that Mr. Black opened each door at most once, and in the end all doors became open.

Input

The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.

The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.

It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.

Output

Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.

Examples

Input


5
0 0 1 0 0


Output


3


Input


4
1 0 0 1


Output


3

Note

In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.

When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.

In the second example when the first two doors were opened, there was open closed door in each of the exit.

With three doors opened Mr. Black was able to use the left exit.
//package javaapplication223;

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a[] = new int[n];
        for(int i=0; i<n; i++) {
            a[i] = sc.nextInt();
        }
        int k = n-1;
        int zero = 0;
        int one = 0;
        for(int i=n-1; i>=0; i--) {
            if(a[i] == 0) {
                zero = 1;
            } else one = 1;
            if(zero == 1 && one == 1) {
                k = i+1;
                break;
            }
        }
        System.out.println(k);
    }
    
}



/*//package D;

import java.util.Scanner;

public class NewClass {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long ans = 0;
        for(int i=3; i<=n; i++) {
            ans += (i * (i-1));
        }
        System.out.println(ans);
    }
}

/*package C;

import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.TreeSet;

public class C {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int songs[][] = new int[n][2];
        for(int i=0; i<n; i++) {
            songs[i][0] = sc.nextInt();
            songs[i][1] = sc.nextInt();
        }
        Arrays.sort(songs, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[1] - o2[1];
            }
        });
        PriorityQueue<Integer> lengths = new PriorityQueue<>();
        long pleasure = 0, length = 0;
        for(int i=n-1; i>=0; i--) {
            lengths.add(songs[i][0]);
            length += songs[i][0];
            if(lengths.size() > k) {
                length -= lengths.poll();
            }
            pleasure = Math.max(pleasure, songs[i][1]*length);
        }
        System.out.println(pleasure);
    }
}

/*
declare
i number:= 2+3+5+7+11+13+17+19+23+27;
j number;
status number:= 1;
total number:= 2;
num number := 3;

begin 
	while i<=10;
	loop
		j:=2;
		while j <= sqrt(num)
		loop
			if mod(num, j) = 0 
			then 
				status:=0;
				EXIT;
			end if;
			j:=j+1;
		end loop;
		if status = 1
		then 
			total:=total+num;
			i=:i+1;
		end if;
		status := 1;
		num:=num+1;
	end loop;
	dbmsoutput.putline(total);
end;
declare
total number := 117;
begin
	dbmsoutput.putline(total);
end;


n number;
i number;
temp number;
counter number;
total number;
n:=2;
i:=2;
counter:=0;
temp:=1;

begin
	while counter < 11
	loop
		for i in 2..n/2
		loop
			if mod(n, i) = 0
			then 
				temp:=0;
				exit;
			end if;
		end loop;
		if temp = 1
		then
			counter:=counter+1;
			total:=total+n;
		end if;
		n:=n+1;
	end loop;
	dbms_output.put_line(total); 
end;
		

1,
11,
111,
1111,
11111,
111111,
1111111,
11111111,
111111111,
1111111111,
11111111111,
111111111111,
1111111111111,
11111111111111,
111111111111111,
1111111111111111,
11111111111111111,
111111111111111111,
1111111111111111111
*continous delovery, continous testing, continuous integration.
			
======================================================================
//support for network isolaion, use of snapshots
//testers and developers can share an itrace file to help reproducing and finding issues, allows you to examine the call staxk and local variables.
they are errors not picked up at testing, they are usually found in production.
small sizes, fast iteration.
//degrades application performace
//state preservation
//orchestrator, data protection manager.
//record some actions against the site
//build deploy, build deploy test, build only
analysis
//library
//firewall issues, missing data binding attributes.
//SCVMM tsting lab
//moniters via a "single pane of glass", itegrated with team foudation server
======================================================================
pair programming, delay building anything until required, frequent coummnication, unit testing.
test cases are developed before writing the code to pass them.
code changes to multiple files that are related to the same software change.
too many conflicting ideas
acquiring knowledge around business processes and using it.
proritzed and scheduled with input from the owner.
deliver more features and functionality, higher customer satisfaction, higher application quality, improved developer productivity.
prefers responding to change over following a plan.
processes.
to represent teams with diverse skills working together.
they are templets from which objects are instantiated.

=======================================================================
Graph and pattern workflow.
FTP
branch and merge
the default text editor opens and you enter a commmit message and save
git show master(merge conflicts)
gitignore(status of file)
git diff(review changes).
git branch hotfix(btanch called hotfix).
it's not free.
git branch -d hotfix(to remove a branch called hotfix).
git remote -v (remote with a git command).

//package B;

import java.util.Scanner;

public class B {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt();
            String s = sc.next();
            int leftDistance = 0;
            int rightDistance = 0;
            for(int i=0; i<n; i++)
                if(s.charAt(i) != '>') leftDistance++;
                else break;
            for(int i=n-1; i>=0; i--)
                if(s.charAt(i) != '<') rightDistance++;
                else break;
            System.out.println(Math.min(leftDistance, rightDistance));
        }
    }
}

for sorting an 2d array 
Array.sort(a, new Comparator<int[]>() {
	@override
	
});

//packagepackage C;

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.TreeSet;

public class C {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        HashMap<Integer, Integer> map = new HashMap<>();
        int bea[] = new int[n];
        long max = Long.MIN_VALUE;
        for(int i=0; i<n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            map.put(y, x);
            bea[i] = y;
        }
        Arrays.sort(bea);
        for(int i=0; i<n; i++) {
            int j = i;
            int temp = 0;
            while(j <= (i+k-1) && j < n) {
                int t = map.get(bea[j]);
                temp += t;
                j++;
//                System.out.print(" t"+j+"= " + t);
            }
//            System.out.println();
            max = Math.max(max, (long) (bea[i] * temp));
        }
        System.out.println(max);
    }
}

*/

