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 maxn = 2e5 + 5;
int a[maxn], nex[maxn], b[maxn], pos[maxn];
int f[maxn][21], bin[maxn], c[maxn], mn[maxn][21];
char ans[maxn];
void build(int n) {
  for (int i = 1; i <= n; i++) mn[i][0] = c[i];
  for (int j = 1; j <= bin[n]; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      int t = i + (1 << (j - 1));
      mn[i][j] = min(mn[i][j - 1], mn[t][j - 1]);
    }
  }
}
int query(int x, int y) {
  int j = bin[y - x + 1];
  int k = y - (1 << j) + 1;
  return min(mn[x][j], mn[k][j]);
}
int main() {
  int n, m, q, l, r;
  scanf("%d%d%d", &n, &m, &q);
  bin[0] = -1;
  for (int i = 1; i <= m; i++) bin[i] = bin[i >> 1] + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (i > 1) nex[a[i - 1]] = a[i];
  }
  nex[a[n]] = a[1];
  for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
  for (int i = m; i >= 1; i--) {
    pos[b[i]] = i;
    f[i][0] = pos[nex[b[i]]];
  }
  for (int j = 0; j <= 19; j++) {
    for (int i = 1; i <= m; i++) f[i][j + 1] = f[f[i][j]][j];
  }
  for (int i = 1; i <= m; i++) {
    int len = n - 1, p = i;
    for (int j = 20; j >= 0; j--) {
      if ((1 << j) <= len) len -= 1 << j, p = f[p][j];
    }
    if (!p)
      c[i] = m + 1;
    else
      c[i] = p;
  }
  build(m);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &l, &r);
    if (query(l, r) <= r)
      ans[i] = '1';
    else
      ans[i] = '0';
  }
  printf("%s\n", ans + 1);
  return 0;
}
