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 = 200000 + 5;
int p[MAXN], a[MAXN], lst[MAXN], inxt[MAXN];
int pre[MAXN], c[MAXN], dep[MAXN], nd[MAXN], r[MAXN];
int lj[MAXN], nxt[MAXN], fir[MAXN];
int n, m, q, i, j, cc, bs;
inline void add(int x, int y) {
  lj[++bs] = y;
  nxt[bs] = fir[x];
  fir[x] = bs;
}
inline void read(int &x) {
  cc = getchar();
  while ((cc < 48) || (cc > 57)) cc = getchar();
  x = cc ^ 48;
  cc = getchar();
  while ((cc >= 48) && (cc <= 57)) {
    x = x * 10 + (cc ^ 48);
    cc = getchar();
  }
}
void dfs(int x) {
  int i;
  nd[dep[x]] = x;
  if (dep[x] >= n)
    r[x] = nd[dep[x] - (n - 1)];
  else
    r[x] = m + 1;
  for (i = fir[x]; i; i = nxt[i]) {
    dep[lj[i]] = dep[x] + 1;
    dfs(lj[i]);
  }
}
int main() {
  read(n);
  read(m);
  read(q);
  for (i = 1; i <= n; i += 1) {
    read(p[i]);
    inxt[p[i]] = p[i - 1];
  }
  inxt[p[1]] = p[n];
  for (i = 1; i <= m; i += 1) {
    read(a[i]);
    pre[i] = lst[a[i]];
    lst[a[i]] = i;
    for (j = lst[inxt[a[i]]]; (j) && (c[j] == 0); j = pre[j]) c[j] = i;
  }
  for (i = 1; i <= m; i += 1)
    if (!c[i]) c[i] = m + 1;
  for (i = 1; i <= m; i += 1) add(c[i], i);
  dfs(m + 1);
  for (i = m - 1; i; i -= 1) r[i] = min(r[i + 1], r[i]);
  while (q--) {
    read(i);
    read(j);
    putchar(48 + (r[i] <= j));
  }
  return 0;
}
