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;
int n, m, q;
int a[200001], b[1000001];
int np[200001];
int pre[2000001];
int g[200001][30];
int ng[200001];
int last = 0;
int find(int v) {
  int d = n - 1;
  for (int k = last; k >= 0; --k) {
    if (((1 << k) > d)) continue;
    d -= (1 << k);
    v = g[v][k];
    if (v == 0) return 0;
  }
  return v;
}
int t[10000001];
void build(int v, int tl, int tr) {
  if (tl == tr)
    t[v] = 300000;
  else {
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v] = 300000;
  }
}
void update(int v, int tl, int tr, int pos, int new_val) {
  if (tl == tr) {
    t[v] = new_val;
    return;
  }
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    update(v * 2, tl, tm, pos, new_val);
  else
    update(v * 2 + 1, tm + 1, tr, pos, new_val);
  t[v] = min(t[v * 2], t[v * 2 + 1]);
}
int sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 30000000;
  if (l == tl && r == tr) {
    return t[v];
  }
  int tm = (tl + tr) / 2;
  return min(sum(v * 2, tl, tm, l, min(r, tm)),
             sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
}
vector<int> pre1[200001];
int main() {
  cin >> n >> m >> q;
  int e = 1;
  while (e <= n - 1) {
    e *= 2;
    ++last;
  }
  --last;
  for (int k = 1; k <= n; ++k) {
    cin >> a[k];
    np[a[k]] = a[k - 1];
  }
  np[a[1]] = a[n];
  for (int k = 1; k <= m; ++k) {
    cin >> b[k];
    int neighbour = np[b[k]];
    pre[b[k]] = k;
    pre1[b[k]].push_back(k);
    if (pre[neighbour] == 0) continue;
    for (int i = 0; i < pre1[neighbour].size(); ++i) {
      int nb = pre1[neighbour][i];
      if (g[nb][0] != 0) {
        continue;
      }
      g[nb][0] = k;
    }
    pre1[neighbour].clear();
  }
  for (int k = m; k >= 1; --k) {
    for (int i = 1; i <= 30; ++i) {
      if (g[k][i - 1] == 0) continue;
      g[k][i] = g[g[k][i - 1]][i - 1];
    }
  }
  build(1, 1, m);
  for (int k = 1; k <= m; ++k) {
    ng[k] = find(k);
    if (ng[k] == 0) ng[k] = 3000000;
    update(1, 1, m, k, ng[k]);
  }
  for (int k = 1; k <= q; ++k) {
    int l, r;
    cin >> l >> r;
    if (sum(1, 1, m, l, r) <= r) {
      cout << 1;
    } else
      cout << 0;
  }
  return 0;
}
