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 double eps = 1e-6;
const int dir[8][2] = {-2, -1, -2, 1, 2, -1, 2, 1, -1, 2, -1, -2, 1, 2, 1, -2};
const int maxn = 2e5 + 5;
int n, m, q;
struct node {
  int a, pos;
  bool operator<(const node &b) const { return a < b.a; }
} lisan[maxn];
int tree[maxn * 4];
int w[maxn];
int pre[maxn][27];
int last[maxn];
void build(int root, int a, int b);
void change(int root, int a, int b, int x, int k);
int ask(int root, int a, int b, int x, int y);
int main() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    lisan[i].a = a;
    lisan[i].pos = i;
  }
  sort(lisan + 1, lisan + 1 + n);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &w[i]);
    w[i] = lisan[w[i]].pos;
  }
  memset(pre, -1, sizeof(pre));
  memset(last, -1, sizeof(last));
  for (int i = 1; i <= m; i++) {
    int a = w[i] - 1;
    if (!a) a = n;
    pre[i][0] = last[a];
    last[w[i]] = i;
  }
  for (int i = 1; i <= 20; i++) {
    for (int j = 1; j <= m; j++) {
      if (pre[j][i - 1] != -1) pre[j][i] = pre[pre[j][i - 1]][i - 1];
    }
  }
  build(1, 1, m);
  for (int i = 1; i <= m; i++) {
    int a = i, b = n - 1;
    for (int j = 20; j >= 0 && a != -1; j--) {
      if ((1 << j) <= b) {
        b -= 1 << j;
        a = pre[a][j];
      }
    }
    if (a != -1) change(1, 1, m, i, a);
  }
  for (int i = 1; i <= q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (ask(1, 1, m, a, b) >= a)
      printf("1");
    else
      printf("0");
  }
  return 0;
}
void build(int root, int a, int b) {
  if (a == b) {
    tree[root] = 0;
    return;
  }
  int mid = (a + b) / 2;
  build(root * 2, a, mid);
  build(root * 2 + 1, mid + 1, b);
  tree[root] = 0;
}
void change(int root, int a, int b, int x, int k) {
  if (a == b) {
    tree[root] += k;
    return;
  }
  int mid = (a + b) / 2;
  if (x <= mid) {
    change(root * 2, a, mid, x, k);
  } else {
    change(root * 2 + 1, mid + 1, b, x, k);
  }
  tree[root] = max(tree[root * 2], tree[root * 2 + 1]);
}
int ask(int root, int a, int b, int x, int y) {
  if (x <= a && b <= y) return tree[root];
  if (b < x || a > y) return 0;
  int mid = (a + b) / 2;
  return max(ask(root * 2, a, mid, x, y), ask(root * 2 + 1, mid + 1, b, x, y));
}
