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
//package cf.train;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class PrB2 {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(System.out);

    long[] readLongs() throws IOException {
        String[] s = reader.readLine().split(" ");
        long[] longs = new long[s.length];
        for(int i = 0; i < longs.length; i++) {
            longs[i] = Long.parseLong(s[i]);
        }
        return longs;
    }

    int[] readInts() throws IOException {
        String[] s = reader.readLine().split(" ");
        int[] ints = new int[s.length];
        for(int i = 0; i < ints.length; i++) {
            ints[i] = Integer.parseInt(s[i]);
        }
        return ints;
    }

    int[] tt;
    int tx = 0;
    int readInt() throws IOException {
        if(tt == null || tx >= tt.length) {
            tt = readInts();
            tx = 0;
        }
        return tt[tx++];
    }

    void solve() throws IOException {
        final int n = readInt();
        final int[] a = readInts();
        final int[] pos = new int[n + 1];
        for(int i = 0; i < n; i++) {
            pos[a[i]] = i;
        }
        class SegmentTree {
            int q = 1;
            int[] a;
            {
                while(q <= n) q *= 2;
                a = new int[2 * q];
            }


            void add(int i, int x) {
                for(i += q; i > 0; i >>= 1) {
                    a[i] += x;
                }
            }

            int sum(int l, int r) {
                int ret = 0;
                for(l += q, r += q; l < r; l /= 2, r /= 2) {
                    if(l % 2 == 1) {
                        ret += a[l++];
                    }
                    if(r % 2 == 1) {
                        ret += a[--r];
                    }
                }
                return ret;
            }
        }
        final SegmentTree st = new SegmentTree();
        for(int i = 2; i <= n; i++) {
            if(pos[i] < pos[i - 1]) st.add(i, 1);
        }

        class Fixer {
            void fix(int x) {
                if(x == 1 || x > n) return;
                if(pos[x] > pos[x - 1] && st.sum(x, x + 1) == 1) {
                    st.add(x, -1);
                }
                if(pos[x] < pos[x - 1] && st.sum(x, x + 1) == 0) {
                    st.add(x, 1);
                }
            }
        }
        Fixer fixer = new Fixer();

        int m = readInt();
        for(int i = 0; i < m; i++) {
            int t = readInt();
            int l = readInt();
            int r = readInt();
            if(t == 2) {
                int tmp = a[l - 1];
                a[l - 1] = a[r - 1];
                a[r - 1] = tmp;
                pos[a[l - 1]] = l - 1;
                pos[a[r - 1]] = r - 1;
                fixer.fix(a[l - 1]);
                fixer.fix(a[l - 1] + 1);
                fixer.fix(a[r - 1]);
                fixer.fix(a[r - 1] + 1);
            }
            else {
                writer.println(st.sum(l + 1, r + 1) + 1);
            }
        }
        writer.flush();
    }

    public static void main(String[] args) throws IOException {
        new PrB2().solve();
    }
}
