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 + 10, inf = 1e18;
int n, m, q, hv[maxn], nextt[maxn], a[maxn], p[maxn], dd[maxn], tt[maxn][50],
    b[maxn], it[maxn * 4];
void build(int st, int l, int r) {
  if (l == r) {
    it[st] = b[l];
    return;
  }
  int mid = (l + r) / 2;
  build(st * 2, l, mid);
  build(st * 2 + 1, mid + 1, r);
  it[st] = min(it[st * 2], it[st * 2 + 1]);
}
long long get(int st, int l, int r, int u, int v) {
  if (l > v || r < u) return inf;
  if (u <= l && r <= v) return it[st];
  int mid = (l + r) / 2;
  return min(get(st * 2, l, mid, u, v), get(st * 2 + 1, mid + 1, r, u, v));
}
int tiento(int x, int h) {
  vector<int> gt;
  while (h) gt.push_back(h % 2), h /= 2;
  for (int i = gt.size() - 1; i >= 0; i--)
    if (gt[i]) x = tt[x][i];
  return x;
}
int main() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> hv[i];
  for (int i = 1; i < n; i++) nextt[hv[i]] = hv[i + 1];
  nextt[hv[n]] = hv[1];
  for (int i = 1; i <= m; i++) cin >> a[i];
  p[m + 1] = m + 1;
  for (int i = 1; i <= n; i++) dd[i] = m + 1;
  for (int i = m; i; i--) {
    p[i] = dd[nextt[a[i]]];
    dd[a[i]] = i;
  }
  int k = n, h = 0;
  while (k) k /= 2, h++;
  for (int i = 0; i <= h; i++) tt[m + 1][i] = m + 1;
  for (int i = 1; i <= m; i++) tt[i][0] = p[i];
  for (int j = 1; j <= h; j++)
    for (int i = m; i; i--) tt[i][j] = tt[tt[i][j - 1]][j - 1];
  for (int i = 1; i <= m; i++) b[i] = tiento(i, n - 1);
  build(1, 1, m);
  int u, v;
  for (int i = 1; i <= q; i++) {
    cin >> u >> v;
    if (get(1, 1, m, u, v) <= v)
      cout << 1;
    else
      cout << 0;
  }
}
