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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MAX_N = 200005;
const int LOG_N = 18;
int n, m, q;
int p[MAX_N];
int a[MAX_N];
int posInP[MAX_N];
vector<int> posInA[MAX_N];
int fwd[LOG_N][MAX_N];
int b[MAX_N];
int t[4 * MAX_N];
void build(int i, int l, int r) {
  if (l == r)
    t[i] = b[l];
  else {
    int mid = (l + r) / 2;
    build(2 * i + 1, l, mid);
    build(2 * i + 2, mid + 1, r);
    t[i] = min(t[2 * i + 1], t[2 * i + 2]);
  }
}
int rmq(int i, int l, int r, int lx, int rx) {
  if (l == lx && r == rx)
    return t[i];
  else {
    int mid = (l + r) / 2;
    int res = m;
    if (lx <= mid) res = rmq(2 * i + 1, l, mid, lx, min(rx, mid));
    if (rx > mid)
      res = min(res, rmq(2 * i + 2, mid + 1, r, max(lx, mid + 1), rx));
    return res;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    posInP[p[i]] = i;
  }
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    a[i]--;
    posInA[a[i]].push_back(i);
  }
  for (int i = 0; i < m; i++) {
    int nextInP = (posInP[a[i]] == n - 1) ? p[0] : p[posInP[a[i]] + 1];
    auto it = upper_bound(posInA[nextInP].begin(), posInA[nextInP].end(), i);
    if (it == posInA[nextInP].end())
      fwd[0][i] = m;
    else
      fwd[0][i] = *it;
  }
  for (int j = 0; j < LOG_N; j++) fwd[j][m] = m;
  for (int i = m - 1; i >= 0; i--)
    for (int j = 1; j < LOG_N; j++) fwd[j][i] = fwd[j - 1][fwd[j - 1][i]];
  for (int i = 0; i < m; i++) {
    int v = i;
    for (int j = LOG_N - 1; j >= 0; j--)
      if ((n - 1) & (1 << j)) v = fwd[j][v];
    b[i] = v;
  }
  build(0, 0, m - 1);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    int val = rmq(0, 0, m - 1, l, r);
    cout << (val <= r ? '1' : '0');
  }
  cout << endl;
}
