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 = 2e5 + 5;
int Prev[N];
int par[N][20];
int Rgt[N];
int depth[N];
int tree[4 * N];
void build(int cn, int b, int e) {
  if (b == e) {
    tree[cn] = Rgt[b];
    return;
  }
  int lc = 2 * cn, rc = lc + 1, mid = (b + e) / 2;
  build(lc, b, mid);
  build(rc, mid + 1, e);
  tree[cn] = max(tree[lc], tree[rc]);
}
int query(int cn, int b, int e, int i, int j) {
  if (b > j or e < i) return 0;
  if (b >= i and e <= j) {
    return tree[cn];
  }
  int lc = 2 * cn, rc = lc + 1, mid = (b + e) / 2;
  return max(query(lc, b, mid, i, j), query(rc, mid + 1, e, i, j));
}
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> seq(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> seq[i];
    if (i > 1) Prev[seq[i]] = seq[i - 1];
  }
  Prev[seq[1]] = seq[n];
  vector<int> a(m + 1);
  for (int i = 1; i <= m; i++) cin >> a[i];
  vector<int> last(n + 1, 0);
  for (int i = 1; i <= m; i++) {
    if (last[Prev[a[i]]]) {
      par[i][0] = last[Prev[a[i]]];
      depth[i] = depth[last[Prev[a[i]]]] + 1;
    } else {
      par[i][0] = -1;
      depth[i] = 1;
    }
    for (int k = 1; k < 20; k++) {
      if (par[i][k - 1] != -1) {
        par[i][k] = par[par[i][k - 1]][k - 1];
      } else {
        par[i][k] = -1;
      }
    }
    if (depth[i] >= n) {
      int u = i;
      int X = n - 1;
      for (int k = 19; k >= 0; k--) {
        if ((1 << k) <= X) {
          X -= (1 << k);
          u = par[u][k];
        }
      }
      Rgt[i] = u;
    }
    last[a[i]] = i;
  }
  build(1, 1, m);
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (query(1, 1, m, l, r) >= l) {
      cout << "1";
    } else {
      cout << "0";
    }
  }
  return 0;
}
