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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  int log_n = 32 - __builtin_clz(n);
  vector<int> inv(n);
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p;
    inv[p - 1] = i;
  }
  vector<int> a(m);
  for (auto& i : a) {
    cin >> i;
    i = inv[i - 1];
  }
  vector<int> pos(n, -1);
  vector<pair<int, int>> cand;
  vector<vector<int>> pre(log_n, vector<int>(m, -1));
  for (int i = 0; i < m; ++i) {
    pre[0][i] = a[i] ? pos[a[i] - 1] : -1;
    for (int j = 1; j < log_n && pre[j - 1][i] != -1; ++j) {
      pre[j][i] = pre[j - 1][pre[j - 1][i]];
    }
    int cur = i;
    for (int j = log_n - 1; j >= 0; --j) {
      if (pre[j][cur] != -1) {
        cur = pre[j][cur];
      }
    }
    if (!a[cur]) {
      cand.emplace_back(cur, i);
    }
    pos[a[i]] = i;
  }
  int ndx = 0;
  int last = -1;
  vector<int> res(m, -1);
  sort(cand.begin(), cand.end());
  for (int i = 0; i < m; ++i) {
    if (a[i] == n - 1) {
      last = i;
    }
    while (ndx < int(cand.size()) && cand[ndx].first == i) {
      auto [l, r] = cand[ndx];
      ++ndx;
      if (a[r] == n - 1) {
        res[r] = l;
      } else if (last != -1) {
        int cur = last;
        for (int j = log_n - 1; j >= 0; --j) {
          if (pre[j][cur] != -1 && a[pre[j][cur]] > a[r]) {
            cur = pre[j][cur];
          }
        }
        if (a[cur] == a[r] + 1) {
          res[r] = cur;
        }
      }
    }
    if (i) {
      res[i] = max(res[i], res[i - 1]);
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << (res[r - 1] >= l - 1);
  }
  cout << "\n";
}
