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;
long long MOD = 1e9 + 7;
long long mpow(long long a, long long b) {
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mlog(long long v) {
  long long ans = 20;
  while (v < (((long long)1) << ans)) ans--;
  return ans;
}
void solve() {
  long long n, m, q;
  cin >> n >> m >> q;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long f[n + 1], r[n + 1];
  for (long long i = 0; i < n; i++) {
    f[a[i]] = a[(i + 1) % n];
    r[a[(i + 1) % n]] = a[i];
  }
  long long arr[m];
  for (long long i = 0; i < m; i++) cin >> arr[i];
  vector<long long> v[n + 1];
  long long logn = 20;
  long long dp[m][logn];
  for (long long i = 0; i < m; i++) dp[i][0] = -1;
  for (long long i = 0; i < m; i++) {
    v[arr[i]].push_back(i);
    if (!(long long)v[r[arr[i]]].size()) continue;
    for (long long j = 0; j < (long long)v[r[arr[i]]].size(); j++) {
      dp[v[r[arr[i]]][j]][0] = i;
    }
    v[r[arr[i]]].clear();
  }
  for (long long j = 1; j < logn; j++) {
    for (long long i = 0; i < m; i++) {
      if (dp[i][j - 1] != -1)
        dp[i][j] = dp[dp[i][j - 1]][j - 1];
      else
        dp[i][j] = -1;
    }
  }
  long long ans[m];
  for (long long i = 0; i < m; i++) {
    long long temp = n - 1;
    long long ind = i;
    while (temp > 0) {
      if (!temp) break;
      long long jump = mlog(temp);
      ind = dp[ind][jump];
      if (ind == -1) {
        break;
      }
      temp -= (((long long)1) << jump);
    }
    if (temp > 0) {
      ans[i] = m;
    } else {
      ans[i] = ind;
    }
  }
  for (long long i = m - 2; i >= 0; i--) ans[i] = min(ans[i], ans[i + 1]);
  for (long long i = 0; i < q; i++) {
    long long t1, t2;
    cin >> t1 >> t2;
    t1--, t2--;
    if (ans[t1] <= t2)
      cout << 1;
    else
      cout << 0;
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
