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;
int n, m, q;
int dp[1000005], nr[1000005], last[1000005], rez, prevv[17][200005];
int a[1000005], perm[1000005], poz[1000005], arb[1000005];
void insert(int now, int st, int dr, int poz, int val) {
  int mij = (st + dr) / 2;
  if (st == dr) {
    arb[now] = val;
    return;
  }
  if (poz <= mij)
    insert(now * 2, st, mij, poz, val);
  else
    insert(now * 2 + 1, mij + 1, dr, poz, val);
  arb[now] = max(arb[now * 2], arb[now * 2 + 1]);
}
void query(int now, int st, int dr, int le, int ri) {
  int mij = (st + dr) / 2;
  if (le > dr || ri < st) return;
  if (le <= st && dr <= ri)
    rez = max(rez, arb[now]);
  else {
    query(now * 2, st, mij, le, ri);
    query(now * 2 + 1, mij + 1, dr, le, ri);
  }
}
int anterior(int poz, int mutari) {
  int rez = poz;
  for (int i = 0; i < 17; i++)
    if ((1 << i) & mutari) rez = prevv[i][rez];
  return rez;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  string s;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> perm[i], poz[perm[i]] = i;
  perm[0] = perm[n];
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    int b = poz[a[i]];
    b = last[perm[b - 1]];
    if (b) {
      prevv[0][i] = b;
      nr[i] = nr[b] + 1;
    } else {
      nr[i] = 1;
    }
    last[a[i]] = i;
  }
  for (int i = 1; i < 17; i++)
    for (int j = 1; j <= m; j++) prevv[i][j] = prevv[i - 1][prevv[i - 1][j]];
  for (int i = 1; i <= m; i++) {
    if (nr[i] >= n) dp[i] = anterior(i, n - 1), insert(1, 1, m, i, dp[i]);
  }
  for (int i = 1; i <= q; i++) {
    int c, d;
    cin >> c >> d;
    rez = 0;
    query(1, 1, m, c, d);
    if (rez >= c)
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
