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 par[300005][30];
int a[300005];
int b[300005];
int nxt[3000005];
vector<int> v[3000005];
const int max_node = 3000000;
int tree[max_node];
int A[max_node];
void update(int node, int start, int _end, int idx, int val) {
  if (start == _end) {
    tree[node] = val;
  } else {
    int mid = (start + _end) / 2;
    if (start <= idx && idx <= mid) {
      update(node * 2, start, mid, idx, val);
    } else {
      update(node * 2 + 1, mid + 1, _end, idx, val);
    }
    tree[node] = min(tree[2 * node], tree[2 * node + 1]);
  }
}
int query(int node, int start, int _end, int l, int r) {
  if (r < start || l > _end) {
    return 1e9;
  }
  if (l <= start && _end <= r) {
    return tree[node];
  }
  int mid = (start + _end) / 2;
  int x1 = query(2 * node, start, mid, l, r);
  int x2 = query(2 * node + 1, mid + 1, _end, l, r);
  return min(x1, x2);
}
int main() {
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) {
    nxt[a[i]] = a[i + 1];
  }
  nxt[a[n - 1]] = a[0];
  for (int i = 0; i < m; i++) {
    scanf("%d", &b[i]);
  }
  for (int i = m - 1; i >= 0; i--) {
    int x = b[i];
    if (v[x].size() > 0) {
      v[x].clear();
    }
    v[x].push_back(i);
    int y = nxt[x];
    if (v[y].size() > 0) {
      par[i][0] = v[y][0];
    } else {
      par[i][0] = 1e9;
    }
  }
  for (int j = 1; j < 20; j++) {
    for (int i = 0; i < m; i++) {
      if (par[i][j - 1] != 1e9)
        par[i][j] = par[par[i][j - 1]][j - 1];
      else
        par[i][j] = 1e9;
    }
  }
  for (int i = 0; i < m; i++) {
    int p = i;
    for (int j = 20; j >= 0; j--) {
      if ((n - 1) & (1 << j)) {
        if (p != 1e9) p = par[p][j];
      }
    }
    if (p != 1e9) p++;
    update(1, 1, m, i + 1, p);
  }
  string s;
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int que = query(1, 1, m, x, y);
    if (que <= y) {
      s += '1';
    } else
      s += '0';
  }
  cout << s;
}
