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 INF = 1000000000;
const int N = 200100;
int n, m, q, a[N], dp[N];
void read() {
  static int p[N], id[N];
  scanf("%d%d%d", &m, &n, &q);
  for (int i = 0; i < m; i++) {
    scanf("%d", p + i);
    id[p[i]] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    a[i] = id[a[i]];
  }
}
void solve() {
  static int toOne[N], toM[N], last[N];
  static pair<int, int> s[N];
  fill(toOne, toOne + n + 2, -1);
  fill(toM, toM + n + 2, -1);
  fill(last, last + n + 2, -1);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      toOne[i] = 0;
      last[1] = i;
    } else if (last[a[i] - 1] != -1) {
      toOne[i] = i - last[a[i] - 1] + toOne[last[a[i] - 1]];
      last[a[i]] = i;
    }
  }
  fill(last, last + n + 2, -1);
  for (int i = n - 1; i >= 0; i--)
    if (a[i] == m) {
      toM[i] = 0;
      last[m] = i;
    } else if (last[a[i] + 1] != -1) {
      toM[i] = last[a[i] + 1] - i + toM[last[a[i] + 1]];
      last[a[i]] = i;
    }
  int k = 0;
  for (int i = 0; i < n; i++)
    if (toM[i] != -1) s[k++] = make_pair(a[i], i);
  sort(s, s + k);
  fill(dp, dp + n + 2, -1);
  for (int i = 0; i < n; i++) {
    if (toOne[i] != -1) {
      if (a[i] == m)
        dp[i] = toOne[i];
      else {
        int f = a[i] + 1;
        int l = lower_bound(s, s + k, make_pair(f, -1)) - s;
        int r = upper_bound(s, s + k, make_pair(f, INF)) - s;
        if (s[l].first == f && s[l].second + toM[s[l].second] < i - toOne[i]) {
          while (l < r - 1) {
            int mid = (l + r) / 2;
            if (s[mid].second + toM[s[mid].second] < i - toOne[i])
              l = mid;
            else
              r = mid;
          }
          if (s[l].second + toM[s[l].second] < i - toOne[i])
            dp[i] = i - s[l].second;
        }
      }
    }
    if (i != 0 && dp[i - 1] != -1) {
      if (dp[i] == -1)
        dp[i] = dp[i - 1] + 1;
      else
        dp[i] = min(dp[i], dp[i - 1] + 1);
    }
  }
}
void print() {
  string res;
  int l, r;
  while (q--) {
    scanf("%d%d", &l, &r);
    if (dp[r - 1] != -1 && dp[r - 1] <= r - l)
      res += "1";
    else
      res += "0";
  }
  puts(res.c_str());
}
int main() {
  read();
  solve();
  print();
  return 0;
}
