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 all[400005];
int how[400005];
bool have[400005] = {0};
vector<int> where[400005];
int Next[400005][30];
int ans[400005];
int Pow[35] = {0};
int now = 1;
struct A {
  int l, r, nxl, nxr;
  int small;
} Node[1000005];
void build(int l, int r, int here) {
  Node[here].l = l;
  Node[here].r = r;
  if (l == r) {
    Node[here].small = ans[l];
    return;
  }
  Node[here].nxl = now++;
  Node[here].nxr = now++;
  build(l, (l + r) / 2, Node[here].nxl);
  build((l + r) / 2 + 1, r, Node[here].nxr);
  Node[here].small =
      min(Node[Node[here].nxl].small, Node[Node[here].nxr].small);
}
int Find(int l, int r, int here) {
  if (l == Node[here].l && r == Node[here].r)
    return Node[here].small;
  else if (r <= (Node[here].l + Node[here].r) / 2)
    return Find(l, r, Node[here].nxl);
  else if (l > (Node[here].l + Node[here].r) / 2)
    return Find(l, r, Node[here].nxr);
  else
    return min(Find(l, (Node[here].l + Node[here].r) / 2, Node[here].nxl),
               Find((Node[here].l + Node[here].r) / 2 + 1, r, Node[here].nxr));
}
int main() {
  int N, M, K, Q, i, j, x, y, t, l, r;
  Pow[0] = 1;
  for (i = 1; i < 30; i++) Pow[i] = Pow[i - 1] * 2;
  memset(Next, -1, sizeof(Next));
  scanf("%d %d %d", &N, &M, &K);
  for (i = 0; i < N; i++) {
    scanf("%d", &t);
    all[t - 1] = i;
  }
  for (i = 0; i < M; i++) {
    scanf("%d", &how[i]);
    how[i]--;
    how[i] = all[how[i]];
    where[how[i]].push_back(i);
  }
  for (i = 0; i < M; i++) {
    if (where[(how[i] + 1) % N].empty() || where[(how[i] + 1) % N].back() <= i)
      continue;
    else
      Next[i][0] = *upper_bound(where[(how[i] + 1) % N].begin(),
                                where[(how[i] + 1) % N].end(), i);
  }
  for (i = 1, x = 2; x < M; i++, x *= 2)
    for (j = 0; j < M; j++)
      if (Next[j][i - 1] != -1) Next[j][i] = Next[Next[j][i - 1]][i - 1];
  for (i = 0; i < M; i++) {
    x = N - 1;
    y = i;
    for (j = 29; j >= 0; j--) {
      if (Pow[j] <= x) {
        y = Next[y][j];
        if (y == -1) break;
        x -= Pow[j];
      }
    }
    ans[i] = y;
    if (ans[i] == -1) ans[i] = 2147483647;
  }
  build(0, M - 1, 0);
  while (K--) {
    scanf("%d %d", &l, &r);
    l--;
    r--;
    printf("%d", Find(l, r, 0) <= r);
  }
  return 0;
}
