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 + 10;
const long long mod = 1000000007;
int n, m, q, p[200005], a[200009], val[200009], sparse[200009][20], lg[200009],
    pos[200009], vis[200009];
vector<int> id[200009];
vector<int> v[200009];
void sp_cal() {
  for (int j = 0; j <= lg[m]; j++)
    for (int i = 0; i + (1 << j) - 1 < m; i++)
      if (j == 0)
        sparse[i][j] = val[i];
      else
        sparse[i][j] = min(sparse[i][j - 1], sparse[i + (1 << (j - 1))][j - 1]);
}
int solve(int l, int r) {
  int len = lg[r - l + 1];
  return min(sparse[l][len], sparse[r - (1 << len) + 1][len]);
}
vector<int> nodes;
void dfs(int node) {
  vis[node]++;
  nodes.push_back(node);
  if (nodes.size() >= n) {
    int z = nodes.size();
    val[nodes.back()] = nodes[z - n];
  }
  for (auto i : v[node])
    if (!vis[i]) dfs(i);
  nodes.pop_back();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  lg[1] = 0;
  for (int i = 2; i <= 2e5 + 5; i++) lg[i] = lg[i / 2] + 1;
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    pos[p[i]] = i;
  }
  for (int i = 0; i < m; i++) cin >> a[i], id[a[i]].push_back(i);
  for (int i = 0; i < m; i++) {
    val[i] = 1e9;
    int my_next = p[(pos[a[i]] + 1) % n];
    int ind = upper_bound(id[my_next].begin(), id[my_next].end(), i) -
              id[my_next].begin();
    if (ind < id[my_next].size()) {
      int par = id[my_next][ind];
      v[par].push_back(i);
    }
  }
  memset(vis, 0, sizeof vis);
  for (int i = m - 1; i >= 0; i--)
    if (!vis[i]) dfs(i);
  sp_cal();
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << (solve(l, r) <= r ? 1 : 0);
  }
  return 0;
}
