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;
const long long maxn = 2e5 + 5;
const long long lg = 21;
const long long mod = 1e9 + 7;
long long z[maxn];
long long nxt[maxn];
long long up[maxn][lg];
long long t[4 * maxn];
void build(long long v, long long tl, long long tr) {
  if (tl == tr)
    t[v] = z[tl];
  else {
    long long tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v] = min(t[v * 2], t[v * 2 + 1]);
  }
}
long long get(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l > r) return 2e9;
  if (tl == l && tr == r) return t[v];
  long long tm = (tl + tr) / 2;
  return min(get(v * 2, tl, tm, l, min(r, tm)),
             get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  long long n, m, q;
  cin >> n >> m >> q;
  long long p[n];
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (long long i = 0; i < n - 1; i++) {
    nxt[p[i]] = p[i + 1];
  }
  nxt[p[n - 1]] = p[0];
  long long a[m];
  long long next_[m];
  for (long long i = 0; i < m; i++) cin >> a[i];
  long long lst[n];
  for (long long i = 0; i < n; i++) {
    lst[i] = m;
  }
  for (long long i = 0; i < lg; i++) up[m][i] = m;
  for (long long i = m - 1; i >= 0; i--) {
    a[i]--;
    next_[i] = lst[nxt[a[i]]];
    up[i][0] = next_[i];
    lst[a[i]] = i;
    for (long long j = 1; j < lg; j++) {
      up[i][j] = up[up[i][j - 1]][j - 1];
    }
    long long now = n - 1;
    long long x = i;
    for (long long j = 20; j >= 0; j--) {
      if ((1 << j) <= now) {
        now -= (1 << j);
        x = up[x][j];
      }
    }
    z[i] = x;
  }
  build(1, 0, m - 1);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    long long pep = get(1, 0, m - 1, l, r);
    if (pep <= r && pep >= l) {
      cout << '1';
    } else
      cout << '0';
  }
  return 0;
}
