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 nmax = 200005;
vector<int> l[nmax];
int p[nmax], inv[nmax], wh[nmax], v[nmax], cons[nmax], dp1[nmax], dp2[nmax],
    lst[nmax];
int ok;
char ans[nmax];
int arb[4 * nmax];
int val, x;
int n, m, q, i, j, poz, nr, st, dr, mn, po;
void update(int nod, int l, int r) {
  arb[nod] = min(arb[nod], val);
  if (l == r) return;
  int mi = (l + r) / 2;
  if (poz <= mi)
    update(2 * nod, l, mi);
  else
    update(2 * nod + 1, mi + 1, r);
}
void query(int nod, int l, int r) {
  if (st <= l && r <= dr) {
    mn = min(mn, arb[nod]);
    return;
  }
  int mi = (l + r) / 2;
  if (st <= mi) query(2 * nod, l, mi);
  if (mi < dr) query(2 * nod + 1, mi + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> q;
  for (i = 1; i <= n; i++) cin >> p[i];
  for (i = 1; i <= n; i++) inv[p[i]] = i;
  for (i = 1; i <= m; i++) {
    cin >> v[i];
    wh[i] = inv[v[i]];
  }
  for (i = 1; i <= m; i++) {
    if (wh[i] == wh[i - 1] + 1)
      cons[i] = cons[i - 1] + 1;
    else
      cons[i] = 1;
  }
  dp1[0] = 0;
  dp2[0] = (1 << 30);
  for (i = 1; i <= m; i++) {
    dp1[i] = 0;
    x = wh[i];
    if (wh[i] == 1)
      dp1[i] = i;
    else {
      dp1[i] = max(dp1[i], dp1[lst[x - 1]]);
    }
    lst[x] = i;
  }
  for (i = 1; i <= n; i++) lst[i] = 0;
  for (i = m; i >= 1; i--) {
    dp2[i] = (1 << 30);
    x = wh[i];
    if (wh[i] == n)
      dp2[i] = i;
    else {
      dp2[i] = min(dp2[i], dp2[lst[x + 1]]);
    }
    lst[x] = i;
  }
  for (i = 1; i <= m; i++) {
    l[wh[i]].push_back(i);
  }
  for (i = 1; i <= max(4 * m, 50); i++) arb[i] = (1 << 30);
  for (i = 1; i <= n; i++) {
    if (i == 1) {
      for (j = 0; j < l[i].size(); j++) {
        poz = l[i][j];
        val = dp2[l[i][j]];
        update(1, 1, m);
      }
    } else {
      po = 0;
      for (j = 0; j < l[i].size(); j++) {
        while (po < l[i - 1].size() && dp1[l[i - 1][po]] <= dp2[l[i][j]]) po++;
        if (po < l[i - 1].size()) {
          poz = l[i][j];
          val = l[i - 1][po];
          update(1, 1, m);
        }
      }
    }
  }
  for (i = 1; i <= q; i++) {
    cin >> st >> dr;
    ok = 1;
    mn = (1 << 30);
    query(1, 1, m);
    ok = (mn <= dr);
    ans[nr++] = ok + '0';
  }
  cout << ans;
  return 0;
}
