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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int64_t N = 202000;
vector<int64_t> g[N];
int64_t f[N][22];
bool in[N];
void dfs(int64_t v, int64_t p) {
  f[v][0] = p;
  for (int64_t i = 1; i < 22; i++) f[v][i] = f[f[v][i - 1]][i - 1];
  for (int64_t x : g[v]) dfs(x, v);
}
int64_t up(int64_t v, int64_t n) {
  for (int64_t i = 21; i >= 0; i--) {
    if ((1LL << i) & n) {
      v = f[v][i];
      n ^= (1LL << i);
    }
  }
  assert(n == 0);
  return v;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t n, m, q;
  cin >> n >> m >> q;
  vector<int64_t> p(n), ar(m), pos(n + 1);
  for (int64_t i = 0; i < n; i++) {
    cin >> p[i];
    pos[p[i]] = i;
  }
  set<int64_t> loc[n + 1];
  for (int64_t i = 0; i < m; i++) {
    cin >> ar[i];
    loc[ar[i]].insert(i);
  }
  for (int64_t i = 0; i < m; i++) {
    int64_t nxt = p[(pos[ar[i]] + 1) % n];
    auto it = loc[nxt].upper_bound(i);
    if (it != loc[nxt].end()) {
      g[*it].push_back(i);
      in[i] = 1;
    }
  }
  for (int64_t i = 0; i < m; i++)
    if (!in[i]) dfs(i, i);
  vector<int64_t> ans(m, -1);
  for (int64_t i = 0; i < m; i++) {
    int64_t d = up(i, n - 1);
    if (ar[d] == p[(pos[ar[i]] - 1 + n) % n]) ans[d] = max(ans[d], i);
  }
  for (int64_t i = 1; i < m; i++) ans[i] = max(ans[i], ans[i - 1]);
  while (q--) {
    int64_t l, r;
    cin >> l >> r;
    if (ans[r - 1] < l - 1)
      cout << 0;
    else
      cout << 1;
  }
}
