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 GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long n, m, k;
const long long maxn = 3e5 + 1;
const long long inf = 1e10;
const long long maxlog = 20;
long long P[maxn];
long long A[maxn];
long long C[maxn];
long long NXT[maxn];
long long nxt[maxn];
long long PREV[maxn];
long long prev[maxn];
long long ANS[maxn];
vector<vector<long long> > PSL;
long long up[maxn][maxlog];
long long st2[maxn];
long long getk(long long x, long long k) {
  long long i = x;
  long long q;
  for (q = maxlog - 1; q >= 0; q--) {
    if (st2[q] <= k) {
      i = up[i][q];
      k -= st2[q];
    }
    if (i == -1) {
      return inf;
    }
  }
  return i;
}
int main() {
  long long q, w, e, l, r;
  long long a, b;
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  st2[0] = 1;
  for (q = 1; q < maxn; q++) {
    st2[q] = st2[q - 1] * 2;
  }
  for (q = 0; q < n; q++) {
    cin >> P[q];
    P[q]--;
  }
  for (q = 1; q < n; q++) {
    NXT[P[q - 1]] = P[q];
    PREV[P[q]] = P[q - 1];
  }
  NXT[P[n - 1]] = P[0];
  PREV[P[0]] = P[n - 1];
  for (q = 0; q < m; q++) {
    cin >> A[q];
    A[q]--;
  }
  for (q = 0; q < n; q++) {
    C[q] = -1;
  }
  for (q = 0; q < n; q++) {
    C[q] = -1;
  }
  for (q = m - 1; q >= 0; q--) {
    nxt[q] = C[NXT[A[q]]];
    C[A[q]] = q;
  }
  for (q = 0; q < m; q++) {
    for (w = 0; w < maxlog; w++) {
      up[q][w] = -1;
    }
  }
  for (q = m - 1; q >= 0; q--) {
    up[q][0] = nxt[q];
    for (w = 1; w < maxlog; w++) {
      if (up[q][w - 1] == -1) {
        break;
      }
      up[q][w] = up[up[q][w - 1]][w - 1];
    }
  }
  for (q = m - 1; q >= 0; q--) {
    ANS[q] = getk(q, n - 1);
  }
  for (q = m - 2; q >= 0; q--) {
    ANS[q] = min(ANS[q + 1], ANS[q]);
  }
  string ans = "";
  for (e = 0; e < k; e++) {
    cin >> l >> r;
    l--;
    r--;
    if (ANS[l] <= r) {
      ans += '1';
    } else {
      ans += '0';
    }
  }
  cout << ans;
  return 0;
}
