The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!

There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a1, a2, ..., an of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i1 < i2 < ... < ik, that ai1 = x, ai2 = x + 1, ..., aik - 1 = y - 1, aik = y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1, 2, 3, ..., n.

If we can't shave beavers from x to y in one session, then we can split these beavers into groups [x, p1], [p1 + 1, p2], ..., [pm + 1, y] (x ≤ p1 < p2 < ... < pm < y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m + 1 working sessions to shave beavers from x to y.

All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types: 

  * what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive? 
  * two beavers on positions x and y (the beavers ax and ay) swapped. 



You can assume that any beaver can be shaved any number of times.

Input

The first line contains integer n — the total number of beavers, 2 ≤ n. The second line contains n space-separated integers — the initial beaver permutation.

The third line contains integer q — the number of queries, 1 ≤ q ≤ 105. The next q lines contain the queries. Each query i looks as pi xi yi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1 ≤ xi < yi ≤ n.

  * to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem B1); 
  * to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems B1+B2). 



Note that the number of queries q is limited 1 ≤ q ≤ 105 in both subproblem B1 and subproblem B2.

Output

For each query with pi = 1, print the minimum number of Beavershave 5000 sessions.

Examples

Input

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5


Output

2
1
3
5
import java.util.*;
import java.io.*;
import java.math.*;

public class B1 {
    public static StreamTokenizer in = new StreamTokenizer(System.in);
    public static boolean bg = false;

    public static void main(String[] args) throws Exception {
        int n1 = ni();
        int[] l1 = intl(n1);
        for (int i=0;i<n1;i++){
            l1[i]--;
        }
        int[] id = new int[n1];
        for (int i=0;i<n1;i++){
            id[l1[i]] = i;
        }
        
        
        int[] inv = new int[n1];
        Fenwick fw1 = new Fenwick(n1);
        for (int i=0;i<n1-1;i++){
            if (id[i] > id[i+1]){
                fw1.add(i);
                inv[i] = 1;
            }
        }
        if (bg) pn(fw1);
        
        int n2 = ni();
        StringBuilder fin = new StringBuilder();
        for (int i=0;i<n2;i++){
            int qu = ni();
            int w1 = ni()-1;
            int w2 = ni()-1;
            if (qu == 1){
                int val = (int)fw1.get(w2-1);
                if (bg) pn(val);
                if (w1>0){
                    val -= (int)fw1.get(w1-1);
                }
                if (bg) pn(val);
                fin.append((val+1)+"\n");
            }
            else {
                int v1 = l1[w1];
                int v2 = l1[w2];
                l1[w2] = v1;
                l1[w1] = v2;
                
                
                int e1 = id[v1];
                int e2 = id[v2];
                id[v1] = e2;
                id[v2] = e1;
                
                if (v1-1>=0){
                    int qqq = v1-1;
                    int nu = 0;
                    if (id[qqq]>id[qqq+1]) nu = 1;
                    int diff = nu - inv[qqq];
                    fw1.add(qqq, diff);
                    inv[qqq] = nu;
                }
                if (v1<n1-1){
                    int qqq = v1;
                    int nu = 0;
                    if (id[qqq]>id[qqq+1]) nu = 1;
                    int diff = nu - inv[qqq];
                    fw1.add(qqq, diff);
                    inv[qqq] = nu;
                }
                if (v2-1>=0){
                    int qqq = v2-1;
                    int nu = 0;
                    if (id[qqq]>id[qqq+1]) nu = 1;
                    int diff = nu - inv[qqq];
                    fw1.add(qqq, diff);
                    inv[qqq] = nu;
                }
                if (v2<n1-1){
                    int qqq = v2;
                    int nu = 0;
                    if (id[qqq]>id[qqq+1]) nu = 1;
                    int diff = nu - inv[qqq];
                    fw1.add(qqq, diff);
                    inv[qqq] = nu;
                }
            }
            if (bg) pn(fw1);
            if (bg) pn(Arrays.toString(inv));
        }
        p(fin);
    }

    private static int ni() throws Exception {
        in.nextToken();
        return (int) in.nval;
    }

    private static long nl() throws Exception {
        in.nextToken();
        return (long) in.nval;
    }

    private static double nd() throws Exception {
        in.nextToken();
        return in.nval;
    }

    private static int[] intl(int n1) throws Exception {
        int[] fin = new int[n1];
        for (int i = 0; i < n1; i++) {
            fin[i] = ni();
        }
        return fin;
    }

    private static void p(Object o1) {
        System.out.print(o1);
    }
    
    private static void pn(Object o1) {
        System.out.println(o1);
    }
    
    
    private static class Fenwick {
        long[] l1 = null;

        public Fenwick(int n1) {
            l1 = new long[n1];
        }

        public static int[] temp = new int[100];
        public static int len = -1;

        public long get(int n1) {
            int k1 = n1 + 1;
            int ptr = 0;
            for (int i = 0;; i++) {
                if (k1 == 0)
                    break;
                if (k1 % 2 == 1) {
                    temp[ptr++] = i;
                }
                k1 = k1 >> 1;
            }
            len = ptr;
            int ptr1 = -1;
            long fin = 0;
            for (int i = len - 1; i >= 0; i--) {
                int cur = temp[i];
                ptr1 += 1 << cur;
                fin += l1[ptr1];
            }
            return fin;
        }

        public void add(int id) {
            add(id, 1);
        }

        public void add(int id, long val) {
            int k1 = id;
            int v = 1;
            for (;;) {
                int idx = k1 * v + v - 1;
                if (idx >= l1.length) {
                    break;
                }
                if (k1 % 2 == 0)
                    l1[idx] += val;
                v = v << 1;
                k1 = k1 >> 1;
            }
        }

        public String toString() {
            long[] fin = new long[l1.length];
            for (int i = 0; i < l1.length; i++)
                fin[i] = get(i);
            return Arrays.toString(fin);
        }
    }
}
