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;
inline int read() {
  char ch = 0;
  int x = 0, flag = 1;
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') flag = -1;
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * flag;
}
struct edge {
  int to, nxt;
} e[440000 * 2];
int num, head[440000];
inline void add(int x, int y) {
  e[++num] = {y, head[x]};
  head[x] = num;
}
int n, m, q, root, cnt, p[440000], g[440000], f[440000], st[440000], lg[440000],
    nxt[440000], w[440000][24];
void dfs(int x, int fa) {
  if (x != root) st[++cnt] = x;
  if (cnt >= n)
    w[x][0] = st[cnt - n + 1];
  else
    w[x][0] = 1e9 + 7;
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int to = e[i].to;
    if (to == fa) continue;
    dfs(to, x);
  }
  cnt--;
}
int query(int l, int r) {
  int k = lg[r - l + 1];
  return min(w[l][k], w[r - (1 << k) + 1][k]);
}
int main() {
  n = read();
  m = read();
  q = read();
  root = m + 1;
  num = -1;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) p[i] = read(), f[i] = root;
  for (int i = 1; i <= n; i++) nxt[p[i]] = p[(i % n) + 1];
  for (int i = 1; i <= m; i++) g[i] = read();
  for (int i = m; i >= 1; i--) {
    int x = g[i];
    add(f[nxt[x]], i), f[x] = i;
  }
  dfs(root, 0);
  for (int i = 1; i <= m; i++)
    lg[i] = lg[i - 1] + ((1 << (lg[i - 1] + 1)) == i);
  for (int k = 1; (1 << k) <= m; k++)
    for (int i = 1; i + (1 << k) - 1 <= m; i++)
      w[i][k] = min(w[i][k - 1], w[i + (1 << (k - 1))][k - 1]);
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    printf("%d", query(l, r) <= r);
  }
  return 0;
}
