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 INF = 0x3f3f3f3f;
vector<int> nidx[200001];
int N, M, Q, p[200001], a[200001], tree[1 << 19], NN, nxt[200001][19],
    pidx[200001];
char ans[200001];
int range_min(int ql, int qr, int l = 1, int r = NN, int idx = 1) {
  if (ql <= l && r <= qr) return tree[idx];
  if (r < ql || qr < l) return INF;
  return min(range_min(ql, qr, l, (l + r) / 2, idx * 2),
             range_min(ql, qr, (l + r) / 2 + 1, r, idx * 2 + 1));
}
int find_idx(int S, vector<int> &is) {
  int l = 0, r = is.size() - 1, m, ret = -1;
  while (l <= r) {
    m = (l + r) / 2;
    if (is[m] <= S)
      l = m + 1;
    else
      ret = is[m], r = m - 1;
  }
  return ret;
}
bool psb(int l, int r) {
  if (r - l + 1 < N) return false;
  return range_min(l, r) <= r;
}
int main() {
  scanf("%d %d %d", &N, &M, &Q);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &p[i]);
    pidx[p[i]] = i;
  }
  for (int i = 1; i <= M; i++) {
    scanf("%d", &a[i]);
    nidx[a[i]].push_back(i);
  }
  memset(tree, 0x3f, sizeof(tree));
  memset(nxt, 0xff, sizeof(nxt));
  for (NN = 1; NN < M; NN <<= 1)
    ;
  for (int i = 1; i <= M; i++)
    nxt[i][0] = find_idx(i, nidx[p[(pidx[a[i]] % N) + 1]]);
  for (int i = 1; i < 19; i++)
    for (int j = 1; j <= M; j++) {
      if (nxt[j][i - 1] < 0) continue;
      nxt[j][i] = nxt[nxt[j][i - 1]][i - 1];
    }
  for (int i = 1; i <= M; i++) {
    int cur = i, copy = N - 1;
    for (int j = 0; copy && cur > 0; j++, copy >>= 1)
      if (copy & 1) cur = nxt[cur][j];
    if (cur < 0) continue;
    tree[NN + i - 1] = cur;
  }
  for (int i = NN - 1; i > 0; i--) tree[i] = min(tree[i * 2], tree[i * 2 + 1]);
  for (int i = 0, l, r; i < Q; i++) {
    scanf("%d %d", &l, &r);
    ans[i] = '0' + psb(l, r);
  }
  printf("%s\n", ans);
  return 0;
}
