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 = 2e5 + 5;
struct seg_tree {
  vector<int> seg;
  int n;
  seg_tree(int x, vector<int>& v) : n(x) {
    seg.resize(4 * n + 5);
    build(v);
  }
  void build(vector<int>& v, int l = 0, int r = -1, int i = 0) {
    if (r == -1) r += n;
    if (l == r) return void(seg[i] = v[l]);
    int m = (l + r) >> 1;
    build(v, l, m, i * 2 + 1);
    build(v, m + 1, r, i * 2 + 2);
    seg[i] = min(seg[i * 2 + 1], seg[i * 2 + 2]);
  }
  int query(int x, int y, int l = 0, int r = -1, int i = 0) {
    if (r == -1) r += n;
    if (r < x || l > y) return N;
    if (l >= x && r <= y) return seg[i];
    int m = (l + r) >> 1;
    return min(query(x, y, l, m, i * 2 + 1), query(x, y, m + 1, r, i * 2 + 2));
  }
};
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> v(n + 1, 0), pos(n + 1, 0);
  for (int j = 1; j <= n; j++) cin >> v[j], pos[v[j]] = j;
  v[0] = v[n];
  vector<int> a(m + 1), prv(m + 1), just(n + 1);
  for (int j = 1; j <= m; j++)
    cin >> a[j], prv[j] = just[v[pos[a[j]] - 1]], just[a[j]] = j;
  vector<vector<int> > dp(21, vector<int>(m + 1, 0));
  for (int j = 1; j <= m; j++) {
    dp[0][j] = prv[j];
    for (int k = 1; k <= 20; k++) dp[k][j] = dp[k - 1][dp[k - 1][j]];
  }
  vector<int> mn_ind(m + 1, N);
  for (int j = 1; j <= m; j++) {
    int x = n - 1;
    int cur = j;
    for (int k = 20; k >= 0; k--)
      if (x & (1 << k)) {
        x -= (1 << k);
        cur = dp[k][cur];
      }
    if (cur) mn_ind[cur] = min(mn_ind[cur], j);
  }
  vector<int> for_seg(m);
  for (int j = 1; j <= m; j++) for_seg[j - 1] = mn_ind[j] - 1;
  seg_tree s(m, for_seg);
  string ans = "";
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    int mn_poss = s.query(l, r);
    if (mn_poss <= r)
      ans += "1";
    else
      ans += "0";
  }
  cout << ans << endl;
  return 0;
}
