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 = 2e5 + 15;
int n, m, q, t1, t2;
int p[N], np[N], arr[N], f[N], vis[N], cl[N], sparse[19][N], pa[19][N], LG[N];
void bullog() {
  LG[1] = 0;
  for (int i = 2; i <= m; ++i) {
    LG[i] = LG[i - 1] + !(i & (i - 1));
  }
}
void build() {
  if (n == 1) {
    for (int j = 0; j < m; ++j) {
      if (arr[j] != p[0])
        sparse[0][j] = 1e9;
      else
        sparse[0][j] = j;
    }
  } else {
    for (int j = 0; j < m; ++j) {
      if (~cl[j])
        sparse[0][j] = cl[j];
      else
        sparse[0][j] = 1e9;
    }
  }
  for (int i = 1; (1 << i) < m; ++i) {
    for (int j = 0; (1 << i) + j <= m; ++j) {
      sparse[i][j] = min(sparse[i - 1][j], sparse[i - 1][j + (1 << (i - 1))]);
    }
  }
}
int rmq(int l, int r) {
  int sz = r - l + 1;
  int lg = LG[sz];
  return min(sparse[lg][l], sparse[lg][r - (1 << lg) + 1]);
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
  }
  for (int i = 0; i < n; ++i) {
    np[p[i]] = p[i + 1];
  }
  np[p[n - 1]] = p[0];
  for (int j = 0; j < m; ++j) {
    scanf("%d", arr + j);
  }
  if (n != 1) {
    for (int i = m - 1; i > -1; --i) {
      if (f[np[arr[i]]])
        pa[0][i] = f[np[arr[i]]];
      else
        pa[0][i] = -1;
      f[arr[i]] = i;
    }
    for (int i = 1; (1 << i) <= m; ++i) {
      for (int j = 0; j < m; ++j) {
        if (~pa[i - 1][j] && ~pa[i - 1][pa[i - 1][j]])
          pa[i][j] = pa[i - 1][pa[i - 1][j]];
        else
          pa[i][j] = -1;
      }
    }
    bullog();
    for (int j = 0; j < m; ++j) {
      int w = n - 1;
      int cur = j;
      while (w && ~pa[LG[w]][cur]) {
        cur = pa[LG[w]][cur];
        w -= (1 << LG[w]);
      }
      if (w)
        cl[j] = -1;
      else
        cl[j] = cur;
    }
  }
  build();
  while (q--) {
    scanf("%d %d", &t1, &t2);
    --t1, --t2;
    if (rmq(t1, t2) <= t2)
      printf("1");
    else
      printf("0");
  }
  puts("");
}
