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.
#include <bits/stdc++.h>
int main() {
  int nP, nA, nQ;
  scanf("%d %d %d", &nP, &nA, &nQ);
  std::vector<int> perm(nP);
  for (int& item : perm) {
    scanf("%d", &item);
  }
  std::vector<int> src(nA);
  for (int& item : src) {
    scanf("%d", &item);
  }
  const int MAX_POW = 30;
  std::vector<std::vector<int>> nextOnPow(MAX_POW + 1,
                                          std::vector<int>(nA, -1));
  std::vector<int> permPos(nP + 1, -1);
  std::vector<int> srcPos(nP + 1, -1);
  for (int i = 0; i < nP; ++i) {
    permPos[perm[i]] = i;
  }
  for (int i = nA - 1; i >= 0; --i) {
    nextOnPow[0][i] = srcPos[perm[(permPos[src[i]] + 1) % nP]];
    srcPos[src[i]] = i;
  }
  std::vector<int> bestEnd(nA, nA + 1);
  int nSkip = nP - 1;
  for (int i = nA - 1; i >= 0; --i) {
    if (nextOnPow[0][i] != -1) {
      for (int pow = 1; pow <= MAX_POW; ++pow) {
        int curNext = nextOnPow[pow - 1][nextOnPow[pow - 1][i]];
        if (curNext == -1) {
          break;
        }
        nextOnPow[pow][i] = curNext;
      }
    }
    int pos = i;
    for (int j = 0; j <= MAX_POW; ++j) {
      if (nSkip & (1 << j)) {
        pos = nextOnPow[j][pos];
        if (pos == -1) {
          break;
        }
      }
    }
    if (pos != -1 && src[pos] == perm[(permPos[src[i]] + nP - 1) % nP]) {
      bestEnd[i] = pos;
    }
  }
  for (int i = nA - 2; i >= 0; --i) {
    bestEnd[i] = std::min(bestEnd[i], bestEnd[i + 1]);
  }
  for (int iQ = 0; iQ < nQ; ++iQ) {
    int left, right;
    scanf("%d %d", &left, &right);
    left -= 1;
    right -= 1;
    if (bestEnd[left] <= right) {
      printf("1");
    } else {
      printf("0");
    }
  }
  printf("\n");
  return 0;
}
