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 optimize("Ofast")
using namespace std;
int n;
vector<int> rr;
vector<int> used;
void dfs(int v, deque<int> &now, vector<vector<int>> &g) {
  used[v] = 1;
  if (now.size() == n - 1) {
    int t = now.front();
    rr[v] = t;
    if (!g[v].empty()) {
      now.pop_front();
      now.push_back(v);
      for (int i : g[v]) dfs(i, now, g);
      now.pop_back();
      now.push_front(t);
    }
  } else if (!g[v].empty()) {
    now.push_back(v);
    for (int i : g[v]) dfs(i, now, g);
    now.pop_back();
  }
}
vector<int> rmq;
void build(int t, int l, int r) {
  if (l + 1 == r)
    rmq[t] = rr[l];
  else {
    int m = (l + r) / 2;
    build(t * 2 + 1, l, m);
    build(t * 2 + 2, m, r);
    rmq[t] = min(rmq[t * 2 + 1], rmq[t * 2 + 2]);
  }
}
int get(int t, int l, int r, int x, int y) {
  if (y <= l || r <= x) return (int)1e9;
  if (x <= l && r <= y) return rmq[t];
  int m = (l + r) / 2;
  return min(get(t * 2 + 1, l, m, x, y), get(t * 2 + 2, m, r, x, y));
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int m, q;
  cin >> n >> m >> q;
  if (n == 1) {
    for (int i = 0; i < q; ++i) cout << 1;
    cout << endl;
    return 0;
  }
  vector<vector<int>> g(m);
  vector<int> perm(n);
  for (int i = 0; i < n; ++i) cin >> perm[i];
  vector<int> tr(n);
  for (int i = 0; i < n; ++i) tr[perm[i] - 1] = i;
  vector<int> a(m);
  for (int i = 0; i < m; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) a[i] = tr[a[i] - 1];
  vector<int> last(n, -1);
  for (int i = m - 1; i >= 0; --i) {
    last[a[i]] = i;
    if (last[(a[i] + 1) % n] != -1) g[last[(a[i] + 1) % n]].push_back(i);
  }
  rr.resize(m, (int)1e9);
  used.resize(m);
  deque<int> st;
  for (int i = m - 1; i >= 0; --i) {
    if (!used[i]) dfs(i, st, g);
  }
  rmq.resize(m * 4);
  build(0, 0, m);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    if (get(0, 0, m, l - 1, r) <= r - 1)
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
