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 INF = 1e18;
int n, m, q;
int perm[200005];
int ara[200005];
int nxt[200005];
int lst[200005];
int oi[200005];
int rms[200005];
int col[200005];
vector<int> vec[200005];
struct segTree {
  int sz;
  vector<long long> tree;
  segTree(int sz) : sz(sz), tree(4 * sz, INF) {}
  void treeUpdate(int node, int i, int j, int x, long long val) {
    int mid = (i + j) >> 1;
    if (i == j && i == x) {
      tree[node] = val;
      return;
    }
    if (i > x || j < x) return;
    treeUpdate(2 * node, i, mid, x, val);
    treeUpdate(2 * node + 1, mid + 1, j, x, val);
    tree[node] = min(tree[2 * node], tree[2 * node + 1]);
    return;
  }
  void update(int idx, long long v) { treeUpdate(1, 1, sz, idx, v); }
  long long treeQuery(int node, int i, int j, int u, int v) {
    int mid = (i + j) >> 1;
    if (i >= u && j <= v) return tree[node];
    if (i > v || j < u) return INF;
    return min(treeQuery(2 * node, i, mid, u, v),
               treeQuery(2 * node + 1, mid + 1, j, u, v));
  }
  long long query(int l, int r) {
    return (l >= 1 && r <= sz) ? treeQuery(1, 1, sz, l, r) : 0;
  }
};
void dfs(int s, int lev) {
  oi[lev] = s;
  col[s] = 1;
  if (lev >= n) rms[s] = oi[lev - n + 1];
  for (int v : vec[s]) dfs(v, lev + 1);
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; i++) scanf("%d", perm + i);
  for (int i = 1; i <= m; i++) scanf("%d", ara + i);
  for (int i = 1; i < n; i++) nxt[perm[i]] = perm[i + 1];
  nxt[perm[n]] = perm[1];
  for (int i = 1; i <= m; i++) lst[i] = m + 1;
  for (int i = m; i; i--) {
    int x = ara[i];
    if (lst[nxt[x]] <= m) vec[lst[nxt[x]]].push_back(i);
    lst[x] = i;
  }
  for (int i = 1; i <= m; i++) rms[i] = m + 1;
  for (int i = m; i; i--) {
    if (col[i] == 0) dfs(i, 1);
  }
  segTree tree(m);
  for (int i = 1; i <= m; i++) tree.update(i, rms[i]);
  string ans;
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int mn = tree.query(l, r);
    ans += (mn <= r) + '0';
  }
  printf("%s\n", ans.c_str());
  return 0;
}
