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;
const int N = (int)2e5 + 7;
const int INF = (int)1e9 + 7;
int m, n, q;
int v[N];
int urm[N];
vector<int> p[N];
int cur[N];
int nxt[N][19];
int limit[N];
int big[N];
int get(int p, int len) {
  if (len == 0) {
    return p;
  }
  int c = nxt[p][big[len]];
  if (c == INF) {
    return INF;
  } else {
    return get(c, len - (1 << big[len]));
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 1; (1 << i) < N; i++) {
    big[(1 << i)] = i;
  }
  for (int i = 1; i < N; i++) {
    if (big[i] == 0) {
      big[i] = big[i - 1];
    }
  }
  cin >> m >> n >> q;
  for (int i = 1; i <= m; i++) {
    cin >> v[i];
    urm[v[i - 1]] = v[i];
  }
  urm[v[m]] = v[1];
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    p[v[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    int c = urm[v[i]];
    if (cur[c] == p[c].size()) {
      nxt[i][0] = INF;
    } else {
      nxt[i][0] = p[c][cur[c]];
    }
    cur[v[i]]++;
  }
  for (int k = 1; k <= 17; k++) {
    for (int i = 1; i <= n; i++) {
      nxt[i][k] = INF;
      if (nxt[i][k - 1] != INF) {
        int c = nxt[i][k - 1];
        if (nxt[c][k - 1] != INF) {
          nxt[i][k] = nxt[c][k - 1];
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    limit[i] = get(i, m - 1);
  }
  for (int i = n - 1; i >= 1; i--) {
    limit[i] = min(limit[i], limit[i + 1]);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << (r >= limit[l]);
  }
  cout << "\n";
  return 0;
}
