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;
int m;
int q;
int pos_a[201000];
int a[211000];
int b[201000];
int pos_b[201000];
int front[201000];
int st[200100][32];
int st2[201000][32];
int fast(int x, int y) {
  int temp = 1;
  for (int i = 1; i <= y; i++) {
    temp *= x;
  }
  return y;
}
int Log[201100];
void pre() {
  for (int i = 1; i <= n; i++) {
    st[i][0] = front[i];
  }
  for (int j = 1; j <= 27; j++) {
    for (int i = 1; i <= n; i++) {
      st[i][j] = st[st[i][j - 1]][j - 1];
    }
  }
  Log[0] = -1;
  for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
  for (int i = 1; i <= n; i++) {
    int t = m - 1, p = i;
    for (int j = 27; j >= 0; j--) {
      if (t - (1 << j) >= 0) {
        t -= (1 << j);
        p = st[p][j];
      }
    }
    st2[i][0] = p;
  }
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      st2[i][j] = max(st2[i][j - 1], st2[i + (1 << j - 1)][j - 1]);
    }
  }
  return;
}
void make(int l, int r) {
  int len = r - l + 1;
  int t = Log[len];
  if (max(st2[l][t], st2[r - (1 << t) + 1][t]) >= l)
    cout << 1;
  else
    cout << 0;
  return;
}
int main() {
  cin >> m >> n >> q;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a[i]);
    pos_a[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    int t = a[pos_a[b[i]] - 1];
    if (!t) t = a[m];
    front[i] = pos_b[t];
    pos_b[b[i]] = i;
  }
  pre();
  int l, r;
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &l, &r);
    make(l, r);
  }
  cout << endl;
  return 0;
}
