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;
using namespace std;
int n, m, q;
int a[210000];
int nxt[210000][20];
int loc[210000];
int p[210000];
int rp[210000];
int ans[210000];
pair<pair<int, int>, int> queries[210000];
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
    rp[p[i]] = i;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    a[i] = rp[a[i]];
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    r--;
    l--;
    queries[i] = {{r, l}, i};
  }
  memset(loc, -1, sizeof(loc));
  for (int i = 0; i < m; i++) {
    int prev = (a[i] + n - 1) % n;
    nxt[i][0] = loc[prev];
    loc[a[i]] = i;
  }
  for (int lvl = 1; lvl < 20; lvl++) {
    for (int i = 0; i < m; i++) {
      if (nxt[i][lvl - 1] == -1)
        nxt[i][lvl] = -1;
      else
        nxt[i][lvl] = nxt[nxt[i][lvl - 1]][lvl - 1];
    }
  }
  sort(queries, queries + q);
  int cur_query = 0;
  int best = -1;
  for (int i = 0; i < m; i++) {
    int cur = i;
    for (int lvl = 0; lvl < 20; lvl++)
      if ((1 << lvl) & (n - 1)) cur = nxt[cur][lvl];
    best = max(best, cur);
    for (; cur_query < q && queries[cur_query].first.first == i; cur_query++) {
      ans[queries[cur_query].second] =
          (best >= queries[cur_query].first.second);
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d", ans[i]);
  }
}
