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;
typedef class Query {
 public:
  int l, r, id;
  bool operator<(Query b) const { return l < b.l; }
} Query;
int n, m, q;
int *p, *a;
Query *qs;
deque<int> *pool;
deque<int> **vec;
inline void init() {
  scanf("%d%d%d", &n, &m, &q);
  p = new int[(n + 1)];
  a = new int[(m + 1)];
  qs = new Query[(q + 1)];
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    p[x] = i - 1;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", a + i);
    a[i] = p[a[i]];
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &qs[i].l, &qs[i].r);
    qs[i].id = i;
  }
}
char *res;
inline void solve() {
  if (n == 1) {
    res = new char[(q + 4)];
    for (int i = 1; i <= q; i++) {
      res[i] = '1';
    }
    res[q + 1] = 0;
    puts(res + 1);
    return;
  }
  sort(qs + 1, qs + q + 1);
  Query *pq = qs + q;
  res = new char[(q + 4)];
  pool = new deque<int>[(m + 1)];
  vec = new deque<int> *[(n + 1)];
  int R = m + 1;
  for (int i = 0; i < n; i++) {
    vec[i] = NULL;
  }
  for (int i = m; i && pq != qs; i--) {
    int x = a[i], y = (x + 1) % n;
    if (vec[y]) {
      if (vec[x] && vec[x]->size() > vec[y]->size()) {
        for (int i = (signed)vec[y]->size() + 1; i; i--) {
          vec[x]->pop_back();
        }
        while (!vec[y]->empty()) {
          vec[x]->push_back(vec[y]->front());
          vec[y]->pop_front();
        }
        vec[x]->push_back(i);
        vec[y] = NULL;
      } else {
        vec[x] = vec[y];
        vec[y] = NULL;
        vec[x]->push_back(i);
      }
      while ((signed)vec[x]->size() > n) {
        vec[x]->pop_front();
      }
    } else {
      if (vec[x]) {
        vec[x]->pop_back();
        vec[x]->push_back(i);
        goto dispose;
      }
      vec[x] = pool++;
      vec[x]->push_back(i);
    }
    if ((signed)vec[x]->size() == n && vec[x]->front() < R) {
      R = vec[x]->front();
    }
  dispose:
    while (pq != qs && pq->l == i) {
      res[pq->id] = '0' + (pq->r >= R);
      pq--;
    }
  }
  res[q + 1] = 0;
  puts(res + 1);
}
int main() {
  init();
  solve();
  return 0;
}
