Recently Lynyrd and Skynyrd went to a shop where Lynyrd bought a permutation p of length n, and Skynyrd bought an array a of length m, consisting of integers from 1 to n. 

Lynyrd and Skynyrd became bored, so they asked you q queries, each of which has the following form: "does the subsegment of a from the l-th to the r-th positions, inclusive, have a subsequence that is a cyclic shift of p?" Please answer the queries.

A permutation of length n is a sequence of n integers such that each integer from 1 to n appears exactly once in it.

A cyclic shift of a permutation (p_1, p_2, …, p_n) is a permutation (p_i, p_{i + 1}, …, p_{n}, p_1, p_2, …, p_{i - 1}) for some i from 1 to n. For example, a permutation (2, 1, 3) has three distinct cyclic shifts: (2, 1, 3), (1, 3, 2), (3, 2, 1).

A subsequence of a subsegment of array a from the l-th to the r-th positions, inclusive, is a sequence a_{i_1}, a_{i_2}, …, a_{i_k} for some i_1, i_2, …, i_k such that l ≤ i_1 < i_2 < … < i_k ≤ r.

Input

The first line contains three integers n, m, q (1 ≤ n, m, q ≤ 2 ⋅ 10^5) — the length of the permutation p, the length of the array a and the number of queries.

The next line contains n integers from 1 to n, where the i-th of them is the i-th element of the permutation. Each integer from 1 to n appears exactly once.

The next line contains m integers from 1 to n, the i-th of them is the i-th element of the array a.

The next q lines describe queries. The i-th of these lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m), meaning that the i-th query is about the subsegment of the array from the l_i-th to the r_i-th positions, inclusive.

Output

Print a single string of length q, consisting of 0 and 1, the digit on the i-th positions should be 1, if the subsegment of array a from the l_i-th to the r_i-th positions, inclusive, contains a subsequence that is a cyclic shift of p, and 0 otherwise.

Examples

Input

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


Output

110


Input

2 4 3
2 1
1 1 2 2
1 2
2 3
3 4


Output

010

Note

In the first example the segment from the 1-st to the 5-th positions is 1, 2, 3, 1, 2. There is a subsequence 1, 3, 2 that is a cyclic shift of the permutation. The subsegment from the 2-nd to the 6-th positions also contains a subsequence 2, 1, 3 that is equal to the permutation. The subsegment from the 3-rd to the 5-th positions is 3, 1, 2, there is only one subsequence of length 3 (3, 1, 2), but it is not a cyclic shift of the permutation.

In the second example the possible cyclic shifts are 1, 2 and 2, 1. The subsegment from the 1-st to the 2-nd positions is 1, 1, its subsequences are not cyclic shifts of the permutation. The subsegment from the 2-nd to the 3-rd positions is 1, 2, it coincides with the permutation. The subsegment from the 3 to the 4 positions is 2, 2, its subsequences are not cyclic shifts of the permutation.
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scan = new Scanner(new BufferedInputStream(System.in));
    PrintWriter writer = new PrintWriter(new BufferedOutputStream(System.out));

    int n = scan.nextInt();
    int m = scan.nextInt();
    int q = scan.nextInt();

    int[] map = new int[n+1];
    for (int i = 0;i < n;i++) {
      map[scan.nextInt()] = i;
    }

    int[] array = new int[m];
    for (int i = 0;i < m;i++) {
      array[i] = map[scan.nextInt()];
    }

    int[][] howTo = new int[20][m];
    for (int[] i : howTo) {
      Arrays.fill(i, m);
    }

    int[] lastSeen = new int[n];
    Arrays.fill(lastSeen, -1);
    for (int i = m-1;i >= 0;i--) {
      lastSeen[array[i]] = i;

      int myNext = array[i]+1;
      if (myNext == n) myNext = 0;

      if (lastSeen[myNext] != -1) {
        howTo[0][i] = lastSeen[myNext]-i;
      }
    }

    for (int i = 1;i < 20;i++) {
      for (int j = 0;j < m;j++) {
        int reaches = j + howTo[i-1][j];
        if (reaches >= m) {
          continue;
        }

        howTo[i][j] = howTo[i-1][j] + howTo[i-1][reaches];
      }
    }

    int[] res = new int[m];
    for (int start = 0;start < m;start++) {
      int at = start;
      for (int i = 0;(1<<i) < n;i++) {
        if (((n-1)&(1<<i)) != 0) {
          int step = howTo[i][at];
          at += step;
          if (at >= m) {
            break;
          }
        }
      }
      res[start] = at;
    }

    howTo = null;

    Node parent = make(0, m-1, res);

    for (int i = 0;i < q;i++) {
      int x = scan.nextInt()-1;
      int y = scan.nextInt()-1;
      int calc = parent.calc(x, y);
      if (calc <= y) {
        writer.print(1);
      } else {
        writer.print(0);
      }
    }

    writer.println();
    writer.flush();
    writer.close();
  }

  private static Node make(int from, int to, int[] res) {
    if (from == to) {
      Node node = new Node();
      node.from = node.to = from;
      node.minHere = res[from];
      return node;
    }

    Node node = new Node();
    node.left = make(from, (from+to)/2, res);
    node.right = make((from+to)/2+1, to, res);
    node.from = from;
    node.to = to;
    node.minHere = Math.min(node.left.minHere, node.right.minHere);
    return node;
  }

  static class Node {
    Node left, right;
    int from, to;
    int minHere;

    int calc(int x, int y) {
      if (x <= from && y >= to) {
        return minHere;
      } else if(x > to || y < from) {
        return Integer.MAX_VALUE/2;
      } else {
        return Math.min(left.calc(x, y), right.calc(x, y));
      }
    }
  }
}
