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 P[2 << 17], invP[2 << 17];
int A[2 << 17];
vector<int> G[2 << 17];
int nxt[18][2 << 17];
int R[2 << 17];
int main() {
  scanf("%d%d%d", &N, &M, &Q);
  for (int i = 0; i < N; i++) {
    scanf("%d", &P[i]);
    P[i]--;
    invP[P[i]] = i;
  }
  for (int i = 0; i < M; i++) {
    scanf("%d", &A[i]);
    A[i]--;
    G[A[i]].push_back(i);
  }
  for (int i = 0; i <= M; i++) nxt[0][i] = i + 1;
  for (int k = 1; k < 18; k++) {
    for (int i = 0; i <= M; i++) {
      int p = P[(invP[A[i]] + (1 << k - 1)) % N];
      auto it = lower_bound(G[p].begin(), G[p].end(), nxt[k - 1][i]);
      if (it == G[p].end())
        nxt[k][i] = M + 1;
      else
        nxt[k][i] = nxt[k - 1][*it];
    }
  }
  R[M] = M + 1;
  for (int i = M; i--;) {
    R[i] = R[i + 1];
    int L = i;
    int id = invP[A[i]];
    for (int k = 0; k < 18; k++)
      if (N >> k & 1) {
        int p = P[id];
        auto it = lower_bound(G[p].begin(), G[p].end(), L);
        if (it == G[p].end())
          L = M + 1;
        else
          L = nxt[k][*it];
        id = (id + (1 << k)) % N;
        if (L > M) break;
      }
    if (R[i] > L) R[i] = L;
  }
  for (; Q--;) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    printf(R[l] <= r ? "1" : "0");
  }
  puts("");
}
