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;
template <typename T, typename U>
inline void smin(T &first, U second) {
  first = min(first, second);
}
template <typename T, typename U>
inline void smax(T &first, U second) {
  first = max(first, second);
}
const long double eps = 1e-15;
const int maxN = 2 * 100000 + 10, mod = (int)1e9 + 7, inf = (int)1e9 + 1;
int a[maxN], pp[maxN], fp[maxN], last[maxN], ff[maxN], res[maxN], s[maxN];
int n, m, q;
void calc(int c) {
  if (c == 1) {
    for (int i = (int)(0); i < (int)(m); ++i) ff[i] = fp[i];
    return;
  }
  calc(c / 2);
  for (int i = (int)(0); i < (int)(m); ++i) {
    int u = ff[i];
    if (u != -1)
      res[i] = ff[u];
    else
      res[i] = -1;
  }
  if (c % 2)
    for (int i = (int)(0); i < (int)(m); ++i) {
      int u = res[i];
      if (u != -1)
        ff[i] = fp[res[i]];
      else
        ff[i] = -1;
    }
  else
    for (int i = (int)(0); i < (int)(m); ++i) ff[i] = res[i];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = (int)(0); i < (int)(n); ++i) {
    int first;
    cin >> first;
    pp[first] = i + 1;
  }
  for (int i = (int)(0); i < (int)(m); ++i) {
    int first;
    cin >> first;
    a[i] = pp[first];
  }
  memset(last, -1, sizeof(last));
  for (int i = (int)(0); i < (int)(m); ++i) {
    last[a[i]] = i;
    fp[i] = last[((a[i] - 1 + n - 1) % n) + 1];
  }
  for (int i = (int)(0); i < (int)(m); ++i) ff[i] = fp[i];
  if (n - 1) calc(n - 1);
  s[0] = -1;
  for (int i = (int)(0); i < (int)(m); ++i) s[i + 1] = max(s[i], ff[i]);
  for (int i = (int)(0); i < (int)(q); ++i) {
    int first, second;
    cin >> first >> second;
    if (s[second] >= first - 1)
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
