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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B2 {
    
    static StringTokenizer st;
    static BufferedReader in;
    static PrintWriter pw;
    static int[]t;
    static int n;
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        n = nextInt();
        int[]a = new int[n+1];
        int[]ind = new int[n+1];
        for (int i = 1; i <= n; i++) {
            a[i] = nextInt();
            ind[a[i]] = i;
        }
        t = new int[n+1];
        for (int i = 2; i <= n; i++) {
            if (ind[i-1] > ind[i])
                inc(i, 1);
        }
        int q = nextInt();
        while (q > 0) {
            q--;
            int t = nextInt();
            int x = nextInt();
            int y = nextInt();
            if (t==2) {
                int r = a[x];
                a[x] = a[y];
                a[y] = r;
                ind[a[x]] = x;
                ind[a[y]] = y;
                if (Math.abs(a[x]-a[y])==1) {
                    if (a[x]==a[y]+1) {
                        inc(a[x], 1);
                        if (a[x] < n && ind[a[x]+1] >= x && ind[a[x]+1] <= y) {
                            inc(a[x]+1, -1);
                        }
                        if (a[y] > 1 && ind[a[y]-1] >= x && ind[a[y]-1] <= y) {
                            inc(a[y], -1);
                        }
                    }
                    else {
                        inc(a[y], -1);
                        if (a[x] > 1 && ind[a[x]-1] >= x && ind[a[x]-1] <= y) {
                            inc(a[x], 1);
                        }
                        if (a[y] < n && ind[a[y]+1] >= x && ind[a[y]+1] <= y) {
                            inc(a[y]+1, 1);
                        }
                    }
                    continue;
                }
                if (a[x] > 1) {
                    if (ind[a[x]-1] <= y && ind[a[x]-1] >= x)
                        inc(a[x], 1);
                }
                if (a[x] < n) {
                    if (ind[a[x]+1] <= y && ind[a[x]+1] >= x) {
                        inc(a[x]+1, -1);
                    }
                }
                if (a[y] > 1) {
                    if (ind[a[y]-1] <= y && ind[a[y]-1] >= x)
                        inc(a[y], -1);
                }
                if (a[y] < n) {
                    if (ind[a[y]+1] <= y && ind[a[y]+1] >= x) {
                        inc(a[y]+1, 1);
                    }
                }
            }
            else {
                int ans = sum(y)-sum(x)+1;
                pw.println(ans);
            }
        }
        pw.close();
    }
    private static int sum(int ind) {
        int sum = 0;
        for (int i = ind; i >= 1; i = (i & (i-1))) {
            sum += t[i];
        }
        return sum;
    }
    private static void inc(int ind, int delta) {
        for (int i = ind; i <= n; i = (i | (i-1))+1) {
            t[i] += delta;
        }
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }
    
    private static double nextDouble() throws IOException{
        return Double.parseDouble(next());
    }
    
    private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}
