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;
template <typename X, typename Y>
static inline void xmin(X &x, Y y) {
  if (y < x) x = y;
}
template <typename X, typename Y>
static inline void xmax(X &x, Y y) {
  if (x < y) x = y;
}
long long pw(long long x, long long y, long long md) {
  if (!y) return 1;
  if (y & 1) return (x * pw(x, y - 1, md)) % md;
  long long tmp = pw(x, y / 2, md);
  return (tmp * tmp) % md;
}
int n, m, q, l, r, a[200002], b[200002], F[200002], ans[200002], P[200002][18];
vector<int> G[200002];
int get(int x, int d) {
  int cnt = 0;
  while (d) {
    if (d & 1) x = P[x][cnt];
    ++cnt;
    d >>= 1;
  }
  return x;
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    F[a[i - 1]] = a[i];
  }
  F[a[n]] = a[1];
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    G[b[i]].push_back(i);
  }
  for (int i = 1; i <= m; ++i) {
    auto it = lower_bound(G[F[b[i]]].begin(), G[F[b[i]]].end(), i);
    if (it == G[F[b[i]]].end())
      P[i][0] = m + 1;
    else
      P[i][0] = *it;
  }
  for (int j = 1; j < 18; ++j) {
    for (int i = 1; i <= m; ++i) P[i][j] = P[P[i][j - 1]][j - 1];
  }
  for (int i = 1; i <= m; ++i) ans[i] = get(i, n - 1);
  ans[m + 1] = m + 1;
  for (int i = m; i >= 1; --i) {
    if (!ans[i]) ans[i] = m + 1;
    xmin(ans[i], ans[i + 1]);
  }
  while (q--) {
    cin >> l >> r;
    if (ans[l] <= r)
      cout << 1;
    else
      cout << 0;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int TESTS = 1;
  while (TESTS--) solve();
  return 0;
}
