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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int MAXN = 2e5 + 10;
const int LOG = 19;
const int base = 1 << 18;
int tree[base * 2];
int pos[MAXN];
int pa[MAXN][LOG];
int before[MAXN];
string ans;
void prep_pa(int m) {
  int i, j;
  for (j = 1; j < LOG; j++) {
    for (i = 1; i <= m; i++) {
      pa[i][j] = pa[pa[i][j - 1]][j - 1];
    }
  }
}
void prep_tree() {
  int i;
  for (i = base - 1; i >= 1; i--) {
    tree[i] = max(tree[i * 2], tree[i * 2 + 1]);
  }
}
int get(int a, int b) {
  a += base;
  b += base;
  int ans = max(tree[a], tree[b]);
  while (a / 2 != b / 2) {
    if (a % 2 == 0) ans = max(ans, tree[a + 1]);
    if (b % 2 == 1) ans = max(ans, tree[b - 1]);
    a /= 2;
    b /= 2;
  }
  return ans;
}
int main() {
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, q, val, i, j, first, cost, ind, moves, last = 0, res, l, r;
  cin >> n >> m >> q;
  for (i = 1; i <= n; i++) {
    cin >> val;
    if (i == 1) first = val;
    before[val] = last;
    last = val;
  }
  before[first] = last;
  for (i = 1; i <= m; i++) {
    cin >> val;
    pa[i][0] = pos[before[val]];
    pos[val] = i;
  }
  prep_pa(m);
  for (i = 1; i <= m; i++) {
    ind = i;
    moves = n - 1;
    cost = base;
    for (j = LOG - 1; j >= 0; j--) {
      if (cost <= moves) {
        ind = pa[ind][j];
        moves -= cost;
      }
      cost /= 2;
    }
    tree[base + i] = ind;
  }
  prep_tree();
  for (i = 0; i < q; i++) {
    cin >> l >> r;
    res = get(l, r);
    if (res >= l)
      ans.push_back('1');
    else
      ans.push_back('0');
  }
  cout << ans;
}
