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 MAX = 200007;
const int MAXL = 20;
int ancestor[MAX][MAXL];
vector<int> V[MAX];
bool vis[MAX];
int best_so_far[MAX];
int NA[MAX];
int pos[MAX];
int last_pos[MAX];
int A[MAX];
int P[MAX];
int B[MAX];
void dfs(int i, int p) {
  vis[i] = true;
  ancestor[i][0] = p;
  for (int j = 1; j < MAXL; j++) {
    int h = ancestor[i][j - 1];
    if (h == -1) break;
    ancestor[i][j] = ancestor[h][j - 1];
  }
  for (int x : V[i]) {
    dfs(x, i);
  }
}
void solve(int n, int m) {
  for (int i = 0; i < n; i++) {
    cin >> P[i];
    pos[P[i]] = i;
  }
  for (int i = 0; i < m; i++) {
    cin >> A[i];
  }
  for (int i = 0; i <= n; i++) last_pos[i] = -1;
  for (int j = 0; j < m; j++) {
    vis[j] = false;
    V[j].clear();
    best_so_far[j] = -1;
    for (int i = 0; i < MAXL; i++) ancestor[j][i] = -1;
  }
  for (int i = 0; i < m; i++) {
    int j = pos[A[i]];
    int prevj = j == 0 ? n - 1 : j - 1;
    int p = last_pos[P[prevj]];
    B[i] = p;
    if (p != -1) V[p].push_back(i);
    last_pos[A[i]] = i;
  }
  for (int i = 0; i < m; i++)
    if (!vis[i]) dfs(i, -1);
  int dest, p, node;
  for (int i = 0; i < m; i++) {
    dest = n - 1;
    node = i;
    for (int l = MAXL - 1; l >= 0; l--) {
      p = (1 << l);
      int a = ancestor[node][l];
      if (p <= dest && a != -1) {
        dest -= p;
        node = a;
      }
      if (node == -1) break;
    }
    NA[i] = dest == 0 ? node : -1;
  }
  best_so_far[0] = max(-1, NA[0]);
  for (int i = 1; i < m; i++) {
    best_so_far[i] = max(best_so_far[i - 1], NA[i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  solve(n, m);
  int l, r;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    cout << (best_so_far[r - 1] >= l - 1 ? 1 : 0);
  }
  cout << endl;
  return 0;
}
