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(false);
  size_t n, m, q;
  cin >> n >> m >> q;
  vector<uint32_t> perm(n);
  for (size_t i = 0; i < n; ++i) {
    cin >> perm[i];
    perm[i]--;
  }
  vector<uint32_t> a(m);
  for (size_t i = 0; i < m; ++i) {
    cin >> a[i];
    a[i]--;
  }
  vector<uint32_t> next(n);
  next[perm.back()] = perm.front();
  for (size_t i = 0; i < n - 1; ++i) {
    next[perm[i]] = perm[i + 1];
  }
  constexpr static auto IMPOSSIBLE = numeric_limits<size_t>::max();
  size_t log2n = 8 * sizeof(int) - __builtin_clz((int)n);
  vector<vector<size_t>> cycle_ends(log2n);
  for (size_t i = 0; i < log2n; ++i) {
    cycle_ends[i].resize(m);
  }
  vector<size_t> closest_next(n, IMPOSSIBLE);
  for (int i = m - 1; i >= 0; --i) {
    cycle_ends[0][i] = closest_next[next[a[i]]];
    closest_next[a[i]] = i;
  }
  for (size_t i = 1; i < log2n; ++i) {
    for (size_t j = 0; j < m; ++j) {
      auto next = cycle_ends[i - 1][j];
      cycle_ends[i][j] =
          next == IMPOSSIBLE ? IMPOSSIBLE : cycle_ends[i - 1][next];
    }
  }
  vector<size_t> bits;
  for (size_t i = 0; i < log2n; ++i) {
    if (((n - 1) >> i) & 1) {
      bits.emplace_back(i);
    }
  }
  vector<size_t> min_right(m);
  size_t closest = IMPOSSIBLE;
  for (int i = m - 1; i >= 0; --i) {
    size_t next = i;
    for (auto j : bits) {
      next = cycle_ends[j][next];
      if (next == IMPOSSIBLE) {
        break;
      }
    }
    closest = min(closest, next);
    min_right[i] = closest;
  }
  for (size_t i = 0; i < q; ++i) {
    size_t l, r;
    cin >> l >> r;
    cout << (int)(min_right[l - 1] <= r - 1);
  }
  cout << '\n';
  return 0;
}
