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 oo = 1e9;
const int N = 250000 * 2;
const int LOGN = 18;
int n, m, q;
int dp[N][LOGN], p[N], nxt[N], a[N], bk[N], mn[N], seg[4 * N];
void build(int n, int s, int e) {
  if (s == e) {
    seg[n] = mn[s];
    return;
  }
  build(2 * n, s, (s + e) / 2);
  build(2 * n + 1, (s + e) / 2 + 1, e);
  seg[n] = min(seg[2 * n], seg[2 * n + 1]);
}
int get(int n, int s, int e, int l, int r) {
  if (s > r || e < l) return oo;
  if (s >= l && e <= r) return seg[n];
  return min(get(2 * n, s, (s + e) / 2, l, r),
             get(2 * n + 1, (s + e) / 2 + 1, e, l, r));
}
int main() {
  cin >> n >> m >> q;
  memset(dp, -1, sizeof dp);
  memset(nxt, -1, sizeof nxt);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    bk[p[i]] = i;
  }
  for (int i = 1; i <= m; ++i) scanf("%d", a + i);
  for (int i = m; i >= 1; --i) {
    int myNext = bk[a[i]] + 1;
    if (myNext > n) myNext = 1;
    myNext = p[myNext];
    dp[i][0] = nxt[myNext];
    nxt[a[i]] = i;
  }
  for (int k = 1; k < LOGN; ++k) {
    for (int j = 1; j <= m; ++j) {
      if (dp[j][k - 1] == -1) continue;
      dp[j][k] = dp[dp[j][k - 1]][k - 1];
    }
  }
  for (int i = 1; i <= m; ++i) {
    mn[i] = oo;
    int rem = n - 1;
    int go = i;
    for (int k = LOGN - 1; k >= 0; --k) {
      if (rem >= (1 << k)) {
        if (dp[go][k] == -1) break;
        rem -= (1 << k);
        go = dp[go][k];
      }
    }
    if (rem == 0) mn[i] = go;
  }
  build(1, 1, m);
  for (int l, r, i = 0; i < q; ++i) {
    scanf("%d%d", &l, &r);
    int mn = get(1, 1, m, l, r);
    if (mn <= r) {
      printf("1");
    } else {
      printf("0");
    }
  }
  puts("");
  return 0;
}
