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 N = 5e5 + 5;
const int MOD = 1e9 + 7;
int p[N];
int dp[N][30];
int lst[N];
int pos[N];
int t[N];
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = pos[l];
    return;
  }
  int m = (l + r) >> 1;
  build(2 * v, l, m);
  build(2 * v + 1, m + 1, r);
  t[v] = max(t[2 * v], t[2 * v + 1]);
}
long long get(int v, int tl, int tr, int l, int r) {
  if (l <= tl && tr <= r) {
    return t[v];
  }
  int tm = (tl + tr) >> 1;
  if (r <= tm) return get(2 * v, tl, tm, l, r);
  if (l > tm) return get(2 * v + 1, tm + 1, tr, l, r);
  return max(get(2 * v, tl, tm, l, r), get(2 * v + 1, tm + 1, tr, l, r));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, q, nn;
  cin >> n >> m >> q;
  nn = n - 1;
  int f, nx = 0;
  for (int i = 0; i < n; ++i) {
    int tmp;
    cin >> tmp;
    if (!i) f = tmp;
    p[tmp] = nx;
    nx = tmp;
  }
  p[f] = nx;
  fill(lst, lst + N, -1);
  fill(t, t + N, -1);
  for (int i = 0; i < m; ++i) {
    int tmp;
    cin >> tmp;
    lst[tmp] = i;
    dp[i][0] = lst[p[tmp]];
    for (int j = 1; j < 30; ++j) {
      if (dp[i][j - 1] != -1)
        dp[i][j] = dp[dp[i][j - 1]][j - 1];
      else
        dp[i][j] = -1;
    }
    int pz = i;
    for (int j = 29; j >= 0; --j) {
      if (pz == -1) break;
      if ((1 << j) & nn) {
        pz = dp[pz][j];
      }
    }
    pos[i] = pz;
  }
  build(1, 0, m - 1);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    if (get(1, 0, m - 1, l, r) >= l) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  return 0;
}
