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;
typedef vector<int> vi;
typedef pair<int, int> ii;
const int maxn = 200005;
int n, tot = 0, a[maxn], prv[maxn], f[maxn], hd[maxn], t[maxn], nxt[maxn],
       qy[maxn], vis[maxn];
deque<int> dq;
inline void add(int u, int v) {
  t[++tot] = v;
  nxt[tot] = hd[u];
  hd[u] = tot;
}
inline void dfs(int v) {
  vis[v] = 1;
  int u, ln = dq.size();
  if (ln == n) {
    u = dq.front();
    qy[v] = u;
    dq.pop_front();
    for (int i = hd[v]; i; i = nxt[i]) {
      dq.push_back(t[i]);
      dfs(t[i]);
      dq.pop_back();
    }
    dq.push_front(u);
    return;
  }
  for (int i = hd[v]; i; i = nxt[i]) {
    dq.push_back(t[i]);
    dfs(t[i]);
    dq.pop_back();
  }
}
int main() {
  int m, q, t1, t2;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  if (n > 1)
    for (int i = 1; i <= n; i++) prv[a[i]] = a[(i - 2 + n) % n + 1];
  for (int i = 1; i <= m; i++) {
    scanf("%d", &t1);
    f[t1] = i;
    if (f[prv[t1]]) add(f[prv[t1]], i);
  }
  for (int i = 1; i <= m; i++)
    if (!vis[i]) {
      dq.push_back(i);
      dfs(i);
      dq.pop_back();
    }
  for (int i = 1; i <= m; i++) qy[i] = max(qy[i], qy[i - 1]);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &t1, &t2);
    if (t1 <= qy[t2])
      printf("1");
    else
      printf("0");
  }
  puts("");
  return 0;
}
