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;
#pragma warning(disable : 4996)
const int N = 2e5 + 5, L = 20;
int p[N], a[N], pos[N], last[N], dp[N], lb[N], t[N * 4], go[N][L];
bool full[N];
void build(int v, int L, int R) {
  if (L == R) {
    t[v] = full[L] ? lb[L] : -1;
  } else {
    int m = (L + R) >> 1;
    build(v * 2, L, m);
    build(v * 2 + 1, m + 1, R);
    t[v] = max(t[v * 2], t[v * 2 + 1]);
  }
}
int que(int l, int r, int v, int L, int R) {
  if (l > r) return -1;
  if (l == L && r == R) return t[v];
  int m = (L + R) >> 1;
  return max(que(l, min(m, r), v * 2, L, m),
             que(max(m + 1, l), r, v * 2 + 1, m + 1, R));
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    pos[p[i]] = i;
  }
  p[n] = p[0];
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  memset(lb, -1, sizeof(lb));
  memset(last, -1, sizeof(last));
  for (int i = 0; i < m; i++) {
    int pr = p[(pos[a[i]] + n - 1) % n];
    dp[a[i]] = min(n, dp[pr] + 1);
    if (dp[a[i]] == 1) {
      lb[i] = i;
    } else if (dp[pr] == n) {
      int x = last[pr], y = n - 2;
      for (int j = L - 1; j >= 0; j--) {
        if (y >= (1 << j)) {
          y -= (1 << j);
          x = go[x][j];
        }
      }
      lb[i] = x;
    } else {
      lb[i] = lb[last[pr]];
    }
    if (last[pr] != -1) {
      go[i][0] = last[pr];
      for (int j = 1; j < L; j++) go[i][j] = go[go[i][j - 1]][j - 1];
    }
    if (dp[a[i]] == n) full[i] = true;
    last[a[i]] = i;
  }
  build(1, 0, m - 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << (que(l, r, 1, 0, m - 1) >= l);
  }
  return 0;
}
