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 maxn = 2e5 + 3;
int n, m, q, li, ri;
int v[maxn];
int perm[maxn], sig[maxn];
int fa[maxn][23];
int lastap[maxn];
int termino[maxn];
int st[4 * maxn];
void build(int nodo, int l, int r) {
  if (l == r) {
    st[nodo] = termino[l];
    return;
  }
  int md = (l + r) / 2;
  build(2 * nodo + 1, l, md);
  build(2 * nodo + 2, md + 1, r);
  st[nodo] = min(st[2 * nodo + 1], st[2 * nodo + 2]);
  return;
}
int get(int nodo, int l, int r, int sq, int eq) {
  if (l > eq || r < sq) return INT_MAX;
  if (l >= sq && r <= eq) return st[nodo];
  int md = (l + r) / 2;
  return min(get(2 * nodo + 1, l, md, sq, eq),
             get(2 * nodo + 2, md + 1, r, sq, eq));
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &perm[i]);
    sig[perm[i - 1]] = perm[i];
    lastap[i] = m + 1;
  }
  sig[perm[n]] = perm[1];
  for (int i = 1; i <= m + 1; i++) {
    if (i != m + 1) scanf("%d", &v[i]);
    for (int j = 0; j <= 20; j++) fa[i][j] = m + 1;
  }
  for (int i = m; i >= 1; i--) {
    fa[i][0] = lastap[sig[v[i]]];
    for (int j = 1; j <= 20; j++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
    lastap[v[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    int dist = n - 1;
    int cur = i;
    for (int j = 20; j >= 0; j--) {
      if (dist >= (1 << j)) {
        cur = fa[cur][j];
        dist -= (1 << j);
      }
    }
    termino[i] = cur;
  }
  build(0, 1, m);
  while (q--) {
    scanf("%d %d", &li, &ri);
    if (get(0, 1, m, li, ri) <= ri)
      printf("1");
    else
      printf("0");
  }
  return 0;
}
