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 gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
const int maxn = 200005;
int n, m, q;
int stk[maxn], top, p[maxn], a[maxn], nxt[maxn], pos[maxn];
vector<int> to[maxn];
int Log[maxn], Min[20][maxn];
void dfs(int u) {
  stk[++top] = u;
  if (top >= n)
    Min[0][u] = stk[top - n + 1];
  else
    Min[0][u] = m + 1;
  for (int v : to[u]) dfs(v);
  --top;
}
int main() {
  n = gi();
  m = gi();
  q = gi();
  for (int i = 1; i <= n; ++i) p[i] = gi(), nxt[p[i - 1]] = p[i];
  nxt[p[n]] = p[1];
  for (int i = 1; i <= m; ++i) a[i] = gi();
  for (int i = m; i >= 1; --i) {
    if (pos[nxt[a[i]]]) to[pos[nxt[a[i]]]].push_back(i);
    pos[a[i]] = i;
  }
  for (int i = m; i >= 1; --i)
    if (!Min[0][i]) dfs(i);
  for (int i = 2; i <= m; ++i) Log[i] = Log[i >> 1] + 1;
  for (int i = 0; (1 << (i + 1)) <= m; ++i)
    for (int j = 1; j + (1 << (i + 1)) - 1 <= m; ++j)
      Min[i + 1][j] = min(Min[i][j], Min[i][j + (1 << i)]);
  for (int l, r, k, i = 1; i <= q; ++i) {
    l = gi();
    r = gi();
    k = Log[r - l + 1];
    printf("%c", (min(Min[k][l], Min[k][r - (1 << k) + 1]) <= r) + '0');
  }
  return 0;
}
