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 sz = 2e5 + 10;
vector<int> sv[sz], st;
int n, m, q, ar[sz];
void dfs(int v) {
  st.push_back(v);
  if (st.size() >= n) {
    ar[v] = st[st.size() - n];
  } else
    ar[v] = -1;
  for (int a = 0; a < sv[v].size(); a++) {
    int ne = sv[v][a];
    dfs(ne);
  }
  st.pop_back();
}
int dot[sz * 4];
void build(int nu, int l, int r) {
  if (l == r)
    dot[nu] = ar[l];
  else {
    int mid = (l + r) >> 1;
    build(nu * 2, l, mid);
    build(nu * 2 + 1, mid + 1, r);
    dot[nu] = max(dot[nu * 2], dot[nu * 2 + 1]);
  }
}
int lg, rg;
int que(int nu, int l, int r) {
  if (l >= lg and r <= rg)
    return dot[nu];
  else if (l > rg or r < lg)
    return -1;
  else {
    int mid = (l + r) >> 1;
    return max(que(nu * 2, l, mid), que(nu * 2 + 1, mid + 1, r));
  }
}
int main() {
  cin >> n >> m >> q;
  int p[n], pr[n];
  for (int a = 0; a < n; a++) scanf("%d", &p[a]), p[a]--;
  pr[p[0]] = p[n - 1];
  for (int a = 1; a < n; a++) pr[p[a]] = p[a - 1];
  map<int, int> ma;
  for (int a = 0; a < m; a++) {
    int x;
    scanf("%d", &x);
    x--;
    if (ma.find(pr[x]) != ma.end()) {
      sv[ma[pr[x]]].push_back(a);
    }
    ma[x] = a;
  }
  for (int a = 0; a < m; a++) ar[a] = -2;
  for (int a = 0; a < m; a++)
    if (ar[a] == -2) dfs(a);
  build(1, 0, m - 1);
  for (int a = 0; a < q; a++) {
    scanf("%d%d", &lg, &rg);
    lg--, rg--;
    if (que(1, 0, m - 1) >= lg)
      printf("1");
    else
      printf("0");
  }
}
