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;
vector<pair<int, int>> bucket[200001];
int a[200001], p[200001];
int pos[200001];
int up[200001][30];
int step[200001];
int last[200001];
int mv(int pos, int step) {
  int i = 0;
  while (step != 0) {
    if (step % 2 == 1) pos = up[pos][i];
    step /= 2;
    i++;
    if (pos == -1) return -1;
  }
  return pos;
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    pos[p[i]] = i;
  }
  for (int i = 0; i < m; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) last[i] = -1;
  for (int i = 0; i < m; i++) {
    int prev;
    if (pos[a[i]] == 0)
      prev = p[n - 1];
    else
      prev = p[pos[a[i]] - 1];
    up[i][0] = last[prev];
    last[a[i]] = i;
  }
  for (int pow = 1; pow <= 20; pow++) {
    for (int i = 0; i < m; i++) {
      up[i][pow] = -1;
      if (up[i][pow - 1] != -1) {
        int next = up[i][pow - 1];
        if (next != -1) up[i][pow] = up[next][pow - 1];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    bucket[r].push_back({l, i});
  }
  string ans;
  ans.resize(q);
  int max_left = -1;
  for (int i = 0; i < m; i++) {
    int pos = mv(i, n - 1);
    if (pos != -1) max_left = max(max_left, pos);
    for (int j = 0; j < bucket[i].size(); j++)
      if (bucket[i][j].first <= max_left)
        ans[bucket[i][j].second] = '1';
      else
        ans[bucket[i][j].second] = '0';
  }
  printf("%s", ans.c_str());
  return 0;
}
