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>
using namespace std;
const int N = 3e5 + 5;
int n, m, q, f[N], a[N], L[N], R[N][22], g[N], X[N], Z[N][22], LOG[3000000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    f[x] = i;
  }
  for (int i = 1, x; i <= m; i++) {
    cin >> x;
    a[i] = f[x];
  }
  memset(f, 0, sizeof f);
  fill(f, f + N, m + 1);
  a[m + 1] = n;
  for (int i = m; i > 0; i--) {
    R[i][0] = i;
    if (a[i] != n) {
      R[i][1] = f[a[i] + 1];
    } else
      R[i][1] = i, g[i] = f[1];
    f[a[i]] = i;
  }
  for (int i = 0; i <= 20; i++) R[m + 1][i] = m + 1;
  for (int i = m; i > 0; i--) {
    for (int h = 2; h <= 20; h++) {
      R[i][h] = R[R[i][h - 1]][h - 1];
    }
  }
  for (int i = 1; i <= m; i++) {
    int s = i;
    for (int h = 20; h >= 0; h--)
      if (a[R[s][h]] != n) {
        s = R[s][h];
      }
    if (a[s] != n) {
      s = R[s][1];
    }
    if (a[s] != n || s == m + 1) {
      X[i] = m + 1;
      continue;
    }
    if (a[i] == 1)
      X[i] = s;
    else {
      s = g[s];
      if (s == m + 1) {
        X[i] = m + 1;
        continue;
      }
      for (int h = 20; h >= 1; h--)
        if (s + (1 << (h - 1)) <= m + 1 && a[R[s][h]] < a[i]) s = R[s][h];
      if (a[s] != a[i] - 1)
        X[i] = m + 1;
      else
        X[i] = s;
    }
  }
  for (int i = 1; i <= m; i++) {
    Z[i][0] = X[i];
  }
  for (int i = 1; i <= 21; i++)
    for (int j = (1 << (i - 1)); j < (1 << i); j++) LOG[j] = i - 1;
  for (int i = m; i > 0; i--) {
    for (int h = 1; h <= 20; h++)
      if (i + (1 << h) <= m + 1)
        Z[i][h] = min(Z[i][h - 1], Z[i + (1 << (h - 1))][h - 1]);
  }
  for (int i = 0, l, r; i < q; i++) {
    cin >> l >> r;
    if (min(Z[l][LOG[r - l + 1]],
            Z[r - (1 << LOG[r - l + 1]) + 1][LOG[r - l + 1]]) <= r)
      cout << 1;
    else
      cout << 0;
  }
  int ptr = 0;
}
