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;
long long int mul(long long int x, long long int y) {
  long long int ans = 1;
  while (y > 0) {
    if (y & 1) ans = (ans * x) % 1000000007;
    y /= 2;
    x = (x * x) % 1000000007;
  }
  return ans;
};
const long long int N = 200005;
long long int p[N], mm[N], last[N], a[N], ans[N], n;
vector<long long int> v[N];
vector<long long int> st;
void dfs(long long int x) {
  long long int size = v[x].size(), i;
  st.push_back(x);
  long long int ln = st.size();
  if (ln >= n)
    ans[x] = st[ln - n];
  else
    ans[x] = 1000000007;
  for (i = 0; i < size; i++) dfs(v[x][i]);
  st.pop_back();
  return;
}
int main() {
  long long int m, q, i, af;
  scanf("%lld", &n);
  scanf("%lld", &m);
  scanf("%lld", &q);
  for (i = 1; i <= n; i++) scanf("%lld", &p[i]), mm[p[i - 1]] = p[i];
  mm[p[n]] = p[1];
  for (i = 1; i <= m; i++) scanf("%lld", &a[i]);
  for (i = m; i >= 1; i--) {
    af = mm[a[i]];
    if (last[af] != 0) v[last[af]].push_back(i);
    last[a[i]] = i;
  }
  memset(ans, -1, sizeof ans);
  for (i = m; i >= 1; i--)
    if (ans[i] == -1) dfs(i);
  for (i = m - 1; i >= 1; i--) ans[i] = min(ans[i], ans[i + 1]);
  string str;
  while (q--) {
    long long int l, r;
    scanf("%lld", &l);
    scanf("%lld", &r);
    if (ans[l] <= r)
      str += '1';
    else
      str += '0';
  }
  cout << str << "\n";
  return 0;
}
