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 int dr[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dc[] = {0, -1, 0, 1, -1, 1, -1, 1};
const int N = 212345;
int n, m, q, p[N], a[N];
int f[N], g[N], last[N];
vector<int> loc[N];
int maxm[N];
int my_find(int id, int val) {
  int l = 0, r = loc[id].size();
  while (l < r) {
    int mid = l + (r - l) / 2;
    int x = loc[id][mid];
    if (f[x] > val)
      r = mid;
    else
      l = mid + 1;
  }
  if (l == loc[id].size())
    return m + 1;
  else
    return loc[id][l];
}
int tmin[N * 4];
void build(int o, int l, int r) {
  if (l == r)
    tmin[o] = maxm[l];
  else {
    int mid = l + (r - l) / 2;
    build(o * 2, l, mid);
    build(o * 2 + 1, mid + 1, r);
    tmin[o] = min(tmin[o * 2], tmin[o * 2 + 1]);
  }
}
const int INF = int(1e9);
int query(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    return tmin[o];
  } else {
    int mid = l + (r - l) / 2;
    int lmax = INF, rmax = INF;
    if (ql <= mid) lmax = query(o * 2, l, mid, ql, qr);
    if (qr > mid) rmax = query(o * 2 + 1, mid + 1, r, ql, qr);
    return min(lmax, rmax);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    p[x] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    a[i] = p[a[i]];
    loc[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) sort((loc[i]).begin(), (loc[i]).end());
  for (int i = 1; i <= m; i++) {
    last[a[i]] = i;
    if (a[i] == 1)
      f[i] = i;
    else
      f[i] = f[last[a[i] - 1]];
  }
  for (int i = 1; i <= n; i++) last[i] = m + 1;
  g[m + 1] = m + 1;
  for (int i = m; i >= 1; i--) {
    last[a[i]] = i;
    if (a[i] == n)
      g[i] = i;
    else
      g[i] = g[last[a[i] + 1]];
  }
  for (int i = 1; i <= m; i++) {
    if (a[i] == 1)
      maxm[i] = g[i];
    else
      maxm[i] = my_find(a[i] - 1, g[i]);
  }
  build(1, 1, m);
  string ans;
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (query(1, 1, m, l, r) <= r)
      ans.push_back('1');
    else
      ans.push_back('0');
  }
  cout << ans << '\n';
  return 0;
}
