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.*;
import java.util.*;

public class cf331b {
  public static void main(String[] args) {
    FastIO in = new FastIO(), out = in;
    int n = in.nextInt();
    int[] p = new int[n+2];
    p[0] = 0;
    p[n+1] = n+1;
    for(int i=1; i<=n; i++)
      p[i] = in.nextInt();
    int[] inv = new int[n+2];
    for(int i=0; i<p.length; i++)
      inv[p[i]] = i;
    BIT bit = new BIT(n+2);
    for(int i=0; i<=n; i++)
      if(inv[i] > inv[i+1])
        bit.set(i, 1);
    int q = in.nextInt();
    for(int i=0; i<q; i++) {
      int type = in.nextInt();
      int x = in.nextInt();
      int y = in.nextInt();
      if(type == 1) {        
        System.out.println(bit.sum(x,y-1)+1);
      }
      else {
        int v1 = p[x];
        int v2 = p[y];
        
        int tmp = p[x];
        p[x] = p[y];
        p[y] = tmp;
        inv[v1] = y;
        inv[v2] = x;
        if(inv[v1-1] < inv[v1]) bit.set(v1-1, 0);
        else bit.set(v1-1, 1);
        if(inv[v1] < inv[v1+1]) bit.set(v1, 0);
        else bit.set(v1, 1);
        if(inv[v2-1] < inv[v2]) bit.set(v2-1, 0);
        else bit.set(v2-1, 1);
        if(inv[v2] < inv[v2+1]) bit.set(v2, 0);
        else bit.set(v2, 1);
      }
    }
    out.close();
  }
  static class BIT {
    int n;
    int[] v;
    BIT(int N) {
      v = new int[n=N];
    }
    int sum(int x, int y) {
      return sum(y)-sum(x-1);
    }
    int sum(int x) {
        return x<0?0:(v[x]+sum((x&(x+1))-1));
    }
    void set(int p, int x) {
      inc(p,x-sum(p,p));
    }
    void inc(int p, int x) {
      for(; p<n; p|=(p+1)) v[p]+=x;
    }
  }
  static class FastIO extends PrintWriter {
    BufferedReader br;
    StringTokenizer st;

    public FastIO() {
      this(System.in, System.out);
    }

    public FastIO(InputStream in, OutputStream out) {
      super(new BufferedWriter(new OutputStreamWriter(out)));
      br = new BufferedReader(new InputStreamReader(in));
      scanLine();
    }

    public void scanLine() {
      try {
        st = new StringTokenizer(br.readLine().trim());
      } catch (Exception e) {
        throw new RuntimeException(e.getMessage());
      }
    }

    public int numTokens() {
      if (!st.hasMoreTokens()) {
        scanLine();
        return numTokens();
      }
      return st.countTokens();
    }

    public String next() {
      if (!st.hasMoreTokens()) {
        scanLine();
        return next();
      }
      return st.nextToken();
    }

    public double nextDouble() {
      return Double.parseDouble(next());
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }
  }
}
