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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int MR = 2e5 + 10;
int p[MR], a[MR], posp[MR];
vector<int> pos[MR], binDist[MR];
pair<int, int> dp[MR], best[MR];
int getLastB(int x) {
  for (int i = 0; i < (30); ++i)
    if (x & (1 << i)) return i;
  return -1;
}
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]--;
    posp[p[i]] = i;
  }
  for (int i = 0; i < (m); ++i) {
    scanf("%d", &a[i]);
    a[i]--;
    pos[a[i]].push_back(i);
    if (!i) {
      best[0] = dp[0] = make_pair(1, 0);
      continue;
    }
    if (n == 1) {
      best[i] = dp[i] = make_pair(1, i);
      continue;
    }
    best[i] = best[i - 1];
    int wsk = posp[a[i]] - 1;
    if (wsk < 0) wsk += n;
    int prev = p[wsk];
    if (pos[prev].empty()) {
      if (best[i].first == 1) best[i].second = i;
      dp[i] = make_pair(1, i);
      continue;
    }
    int pospop = pos[prev].back();
    {
      int wsk = 0;
      binDist[i].push_back(pospop);
      while (true) {
        wsk++;
        if (binDist[pospop].size() <= wsk - 1) break;
        pospop = binDist[pospop][wsk - 1];
        binDist[i].push_back(pospop);
      }
    }
    pospop = pos[prev].back();
    if (dp[pospop].first < n) {
      dp[i] = dp[pospop];
      dp[i].first++;
    } else {
      int ile = n - 1, akt = i;
      while (ile) {
        akt = binDist[akt][getLastB(ile)];
        ile &= ile - 1;
      }
      dp[i] = make_pair(n, akt);
    }
    best[i] = max(best[i], dp[i]);
  }
  for (int c = 0; c < (q); ++c) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    if (best[r].first == n && best[r].second >= l)
      printf("1");
    else
      printf("0");
  }
  printf("\n");
  return 0;
}
