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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)(1e9 + 1e5);
const long long LINF = (long long)(2e18);
const long long BASE = 1003;
const double EPS = 1e-9;
const int MOD = 1000000007;
const int LOG = 19;
const int MAXN = 200500;
int tnum;
struct state {
  int val;
  int p[LOG];
  int first;
};
struct query {
  int l, r, num;
  bool ans = false;
};
int n, m, k;
int p[MAXN], inv[MAXN];
int a[MAXN];
int last[MAXN];
state dp[MAXN];
query qu[MAXN];
void init() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= k; ++i) {
    scanf("%d%d", &qu[i].l, &qu[i].r);
    qu[i].num = i;
  }
}
void solve() {
  p[0] = p[n];
  for (int i = 1; i <= n; ++i) {
    inv[p[i]] = i;
  }
  sort(qu + 1, qu + k + 1,
       [](const query& lhs, const query& rhs) { return lhs.r < rhs.r; });
  int qu_ptr = 1;
  int best_end = 0;
  for (int i = 1; i <= m; ++i) {
    int pos = inv[a[i]];
    int prev_val = p[pos - 1];
    int prev_pos = last[prev_val];
    if (prev_pos) {
      dp[i].val = dp[prev_pos].val + 1;
      dp[i].p[0] = prev_pos;
      for (int h = 1; h < LOG; ++h) {
        int cur = prev_pos;
        for (int j = 0; j < h; ++j) {
          cur = dp[cur].p[j];
        }
        dp[i].p[h] = cur;
        if (!cur) {
          break;
        }
      }
      dp[i].first = dp[prev_pos].first;
      if (dp[i].val > n) {
        --dp[i].val;
        auto first = i;
        for (int j = 0; j < LOG; ++j) {
          if ((1 << j) & (n - 1)) {
            first = dp[first].p[j];
          }
        }
        dp[i].first = first;
      }
    } else {
      dp[i].first = i;
      dp[i].val = 1;
    }
    last[a[i]] = i;
    if (dp[i].val == n) {
      if (!best_end || (dp[best_end].first < dp[i].first)) {
        best_end = i;
      }
    }
    while (qu_ptr <= k && qu[qu_ptr].r == i) {
      qu[qu_ptr].ans = (best_end && qu[qu_ptr].l <= dp[best_end].first);
      ++qu_ptr;
    }
  }
  sort(qu + 1, qu + k + 1,
       [](const query& lhs, const query& rhs) { return lhs.num < rhs.num; });
}
void print_answer() {
  for (int i = 1; i <= k; ++i) {
    printf("%d", qu[i].ans ? 1 : 0);
  }
}
void precalc() {}
void run() {
  precalc();
  int t = 1;
  for (tnum = 1; tnum <= t; ++tnum) {
    init();
    solve();
    print_answer();
  }
}
int main() {
  srand(90);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  run();
  return 0;
}
