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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = (long long)INF * INF;
const int MAX = 200 * 1000 + 7;
const int LOG = 18;
int p[MAX];
int a[MAX];
int invP[MAX];
vector<int> positions[MAX];
int ptr[MAX];
int par[LOG][MAX];
int ST[LOG][MAX];
int lg[MAX];
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", p + i);
    p[i]--;
    invP[p[i]] = i;
  }
  for (int i = (0); i < (m); ++i) {
    scanf("%d", a + i);
    a[i]--;
    positions[a[i]].push_back(i);
  }
  for (int i = (0); i < (m); ++i) {
    ptr[a[i]]++;
    int pos = invP[a[i]] + 1;
    if (pos == n) {
      pos = 0;
    }
    par[0][i] = ptr[p[pos]] < (int)((positions[p[pos]]).size())
                    ? positions[p[pos]][ptr[p[pos]]]
                    : m;
  }
  par[0][m] = m;
  for (int j = (1); j < (LOG); ++j) {
    for (int i = (0); i < (m + 1); ++i) {
      par[j][i] = par[j - 1][par[j - 1][i]];
    }
  }
  for (int i = (0); i < (m); ++i) {
    ST[0][i] = i;
    for (int j = (LOG)-1; j >= (0); --j) {
      if (((n - 1) >> j) & 1) {
        ST[0][i] = par[j][ST[0][i]];
      }
    }
  }
  for (int j = (1); j < (LOG); ++j) {
    for (int i = (0); i < (m + 1 - (1 << j)); ++i) {
      ST[j][i] = min(ST[j - 1][i], ST[j - 1][i + (1 << (j - 1))]);
    }
  }
  lg[1] = 0;
  for (int i = (2); i < (MAX); ++i) {
    lg[i] = lg[i - 1];
    if ((1 << (lg[i] + 1)) <= i) {
      lg[i]++;
    }
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    int curLg = lg[r - l + 1];
    printf("%d", min(ST[curLg][l], ST[curLg][r - (1 << curLg) + 1]) <= r);
  }
  return 0;
}
