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;
int arr[200005];
int p[200005];
int neto[200005];
int lastseen[200005];
int pad[200005][21];
int answer[200005];
struct st {
  int val;
  st *l, *r;
};
st* root;
st* build(int i, int j) {
  st* n = new st();
  if (i == j) {
    n->val = answer[i];
    return n;
  }
  int mid = (i + j) / 2;
  n->l = build(i, mid);
  n->r = build(mid + 1, j);
  n->val = min(n->l->val, n->r->val);
  return n;
}
int query(st* cur, int i, int j, int x, int y) {
  if (j < x || y < i) return M;
  if (x <= i && j <= y) return cur->val;
  int mid = (i + j) / 2;
  return min(query(cur->l, i, mid, x, y), query(cur->r, 1 + mid, j, x, y));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> Q;
  int x;
  for (int i = 0; i < N; i++) {
    cin >> p[i];
    lastseen[i + 1] = M;
  }
  for (int i = 0; i < 21; i++) pad[M][i] = M;
  for (int i = 0; i < M; i++) {
    cin >> arr[i];
  }
  neto[p[N - 1]] = p[0];
  for (int i = 1; i < N; i++) {
    neto[p[i - 1]] = p[i];
  }
  int s, c;
  for (int i = M - 1; i >= 0; i--) {
    lastseen[arr[i]] = i;
    pad[i][0] = lastseen[neto[arr[i]]];
    for (int j = 1; j < 21; j++) pad[i][j] = pad[pad[i][j - 1]][j - 1];
    s = 0;
    c = i;
    for (int j = 20; j >= 0; j--) {
      if (s + (1 << j) < N) {
        c = pad[c][j];
        s += 1 << j;
      }
    }
    answer[i] = c;
    if (N == 1) answer[i] = M;
    if (N == 1 && p[0] == arr[i]) answer[i] = i;
  }
  root = build(0, M - 1);
  int i, j;
  while (Q--) {
    cin >> i >> j;
    i--;
    j--;
    cout << (query(root, 0, M - 1, i, j) <= j ? 1 : 0);
  }
  return 0;
}
