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 = 200200;
const int K = 20;
const int Inf = 1e9;
int a[N];
int b[N];
int p[N];
int nxt[N];
int lst[N];
int Bit[N];
int Anc[N][K];
int Min[N][K];
vector<int> v[N];
int Get(int i, int kth) {
  int Res = i;
  for (int k = 0; k < K; k++)
    if (kth & (1 << k)) Res = Anc[Res][k];
  return Res;
}
int GetMin(int l, int r) {
  int Jump = Bit[r - l + 1];
  return min(Min[l][Jump], Min[r - (1 << Jump) + 1][Jump]);
}
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  for (int i = 1; i <= m; i++) scanf("%d", a + i), v[a[i]].push_back(i);
  nxt[b[n]] = b[1];
  lst[b[1]] = b[n];
  for (int i = 1; i < n; i++) nxt[b[i]] = b[i + 1];
  for (int i = 2; i <= n; i++) lst[b[i]] = b[i - 1];
  for (int i = 1; i <= m; i++) {
    Anc[i][0] = i;
    auto it = upper_bound(v[nxt[a[i]]].begin(), v[nxt[a[i]]].end(), i);
    if (it != v[nxt[a[i]]].end()) Anc[i][0] = *it;
  }
  for (int k = 1; k < K; k++)
    for (int i = 1; i <= m; i++) Anc[i][k] = Anc[Anc[i][k - 1]][k - 1];
  for (int i = 1; i <= m; i++) {
    p[i] = Inf;
    int g = Get(i, n - 1);
    if (a[g] == lst[a[i]]) p[i] = g;
  }
  for (int i = 1; i < N; i++)
    for (int k = 0; k < K; k++)
      if (i & (1 << k)) Bit[i] = k;
  for (int i = 1; i <= m; i++) Min[i][0] = p[i];
  for (int k = 1; k < K; k++)
    for (int i = 1; i + (1 << k) - 1 <= m; i++)
      Min[i][k] = min(Min[i][k - 1], Min[i + (1 << (k - 1))][k - 1]);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int Ans = GetMin(x, y);
    printf("%d", Ans <= y);
  }
}
