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>
int kx[8] = {+2, +2, -2, -2, +1, +1, -1, -1};
int ky[8] = {+1, -1, +1, -1, +2, -2, +2, -2};
int d9x[9] = {+1, +1, +1, +0, +0, +0, -1, -1, -1};
int d9y[9] = {+1, +0, -1, +1, +0, -1, +1, +0, -1};
int dx4[4] = {+0, +0, +1, -1};
int dy4[4] = {+1, -1, +0, +0};
long long int gcd(unsigned long long int a, unsigned long long int b) {
  return (a == 0) ? b : gcd(b % a, a);
}
long long int lcm(unsigned long long int a, unsigned long long int b) {
  return a * (b / gcd(a, b));
}
const long long INF = 1e18;
using namespace std;
int n, m, q;
int end_loop[200005];
int L[200005 * 4], R[200005 * 4], seg[200005 * 4];
vector<int> adj[200005 + 1];
vector<int> path;
void dfs(int at) {
  path.push_back(at);
  if ((int)path.size() - n > 0) {
    end_loop[at] = path[(int)path.size() - n];
  }
  for (int to : adj[at]) {
    dfs(to);
  }
  path.pop_back();
  return;
}
void build(int at, int l, int r) {
  L[at] = l;
  R[at] = r;
  int M = l + (r - l) / 2;
  if (l == r) {
    seg[at] = ((end_loop[l] == -1) ? 200005 : end_loop[l]);
    return;
  }
  build(at * 2, l, M);
  build(at * 2 + 1, M + 1, r);
  seg[at] = min(seg[at * 2], seg[at * 2 + 1]);
  return;
}
int mrq(int at, int l, int r) {
  if (L[at] >= l && r >= R[at]) {
    return seg[at];
  }
  int M = L[at] + (R[at] - L[at]) / 2;
  int res = 200005 + 1;
  if (l <= M) {
    res = min(res, mrq(at * 2, l, r));
  }
  if (r >= M + 1) {
    res = min(res, mrq(at * 2 + 1, l, r));
  }
  return res;
}
int main() {
  cin >> n >> m >> q;
  int a[n], b[m], match[n];
  int first[n];
  memset(first, -1, sizeof(first));
  memset(end_loop, -1, sizeof(end_loop));
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
    if (i > 0) {
      match[a[i - 1]] = a[i];
    }
  }
  match[a[n - 1]] = a[0];
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
    --b[i];
  }
  for (int i = m - 1; ~i; --i) {
    first[b[i]] = i;
    int par;
    if (first[match[b[i]]] == -1) {
      par = 200005;
    } else {
      par = first[match[b[i]]];
    }
    adj[par].push_back(i);
  }
  dfs(200005);
  build(1, 0, m - 1);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    if (n == 1) {
      cout << 1;
      continue;
    }
    int res = mrq(1, l, r);
    cout << (res <= r);
  }
  cout << endl;
}
