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;
int p[200001];
int s[200001];
int a[200001];
int nxt[21][200001];
int last[200001];
int pr[200001];
int t[1000001];
int n, m, q;
inline int calc(int x) {
  int lim = n - 1;
  for (int i = 20; i >= 0; i--) {
    if ((1 << i) <= lim) {
      x = nxt[i][x];
      lim ^= (1 << i);
    }
  }
  return x;
}
void build(int v = 1, int tl = 1, int tr = m) {
  if (tl == tr) {
    t[v] = pr[tl];
    return;
  }
  int tm = (tl + tr) / 2;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
  t[v] = min(t[v + v], t[v + v + 1]);
}
int get(int l, int r, int v = 1, int tl = 1, int tr = m) {
  if (tr < l || tl > r) return m + 1;
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  return min(get(l, r, v + v, tl, tm), get(l, r, v + v + 1, tm + 1, tr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
  }
  p[n + 1] = p[1];
  for (int i = 1; i <= n; i++) {
    s[p[i]] = p[i + 1];
  }
  for (int i = 1; i <= m; i++) cin >> a[i];
  for (int i = m; i >= 1; i--) {
    int pos = last[s[a[i]]];
    nxt[0][i] = pos;
    for (int j = 1; j <= 20; j++) {
      nxt[j][i] = nxt[j - 1][nxt[j - 1][i]];
    }
    last[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    pr[i] = calc(i);
    if (!pr[i]) pr[i] = m + 1;
  }
  build();
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (get(l, r) <= r)
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
