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;
int n, m, q, a[1000005], b[1000005], yur[22][1000005], git[1000005], c[1000005],
    seg[4 * 1000005];
vector<int> g[1000005];
void bu(int k, int bas, int son) {
  if (bas == son) {
    seg[k] = c[bas];
    return;
  }
  bu((k + k), bas, ((bas + son) / 2));
  bu((k + k + 1), ((bas + son) / 2) + 1, son);
  seg[k] = min(seg[(k + k)], seg[(k + k + 1)]);
}
int qu(int k, int bas, int son, int x, int y) {
  if (bas > y or son < x) return 1000000007;
  if (bas >= x and son <= y) return seg[k];
  return min(qu((k + k), bas, ((bas + son) / 2), x, y),
             qu((k + k + 1), ((bas + son) / 2) + 1, son, x, y));
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) git[a[i]] = a[(i + 1) % n];
  for (int i = 1; i <= m; i++) {
    scanf("%d", b + i);
    g[b[i]].push_back(i);
  }
  for (int i = 1; i <= m; i++) {
    int bul = git[b[i]];
    int ind = lower_bound(g[bul].begin(), g[bul].end(), i) - g[bul].begin();
    if (ind < g[bul].size()) {
      yur[0][i] = g[bul][ind];
    }
  }
  for (int i = 1; i <= 20; i++)
    for (int j = 1; j <= m; j++) yur[i][j] = yur[i - 1][yur[i - 1][j]];
  for (int i = 1; i <= m; i++) {
    int kal = n - 1;
    int ind = i;
    for (int j = 20; j >= 0; j--)
      if ((1ll << j) <= kal) {
        ind = yur[j][ind];
        kal -= (1ll << j);
      }
    if (ind)
      c[i] = ind;
    else
      c[i] = 1000000007;
  }
  bu(1, 1, m);
  while (q--) {
    int bas, son;
    scanf("%d %d", &bas, &son);
    if (qu(1, 1, m, bas, son) <= son)
      putchar('1');
    else
      putchar('0');
  }
  return 0;
}
