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;
template <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
const long long sze = 3e5 + 50, oo = 1e18 + 500, mod = 1e9 + 7;
const long double eps = 1e-9, PI = 2 * acos(0.0);
vector<long long> vertices[sze];
vector<long long> visit(sze, 0);
vector<char> used(sze, false);
vector<long long> arr(sze, 0);
long long n, m, k;
long long A, B;
long long cnt = 0;
const long long N = 1 << 18;
vector<long long> t(N << 1, oo);
void build(vector<long long> &a) {
  for (long long i = 0; i < m; i++) {
    t[i + N] = a[i];
  }
  for (long long i = N - 1; i; i--) {
    t[i] = min(t[i << 1], t[i << 1 | 1]);
  }
}
long long get(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (tl == l && tr == r) {
    return t[v];
  }
  long long md = (tl + tr) >> 1;
  long long ans = oo;
  if (l <= md) {
    ans = min(ans, get(v << 1, tl, md, l, min(r, md)));
    ;
  }
  if (r > md) {
    ans = min(ans, get(v << 1 | 1, md + 1, tr, max(l, md + 1), r));
    ;
  }
  return ans;
}
long long dp[sze][20];
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m >> k;
  vector<long long> p(n), a(m);
  vector<long long> nxt(n + 10);
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    nxt[p[i]] = p[i + 1];
  }
  nxt[p[n - 1]] = p[0];
  vector<long long> ptr(n + 10, 0);
  for (long long i = 0; i < m; i++) {
    cin >> a[i];
    vertices[a[i]].push_back(i);
  }
  for (long long i = 0; i < m; i++) {
    long long v = nxt[a[i]];
    while (ptr[v] < ((long long)(vertices[v].size())) &&
           vertices[v][ptr[v]] < i) {
      ptr[v]++;
    }
    if (ptr[v] < ((long long)(vertices[v].size()))) {
      a[i] = vertices[v][ptr[v]];
    } else {
      a[i] = oo;
    }
  }
  for (long long i = 0; i <= m; i++) {
    for (long long j = 0; j < 20; j++) {
      dp[i][j] = oo;
    }
  }
  for (long long i = m - 1; i >= 0; i--) {
    dp[i][0] = a[i];
    for (long long j = 1; j < 20; j++) {
      if (dp[i][j - 1] != oo)
        dp[i][j] = dp[dp[i][j - 1]][j - 1];
      else
        break;
    }
  }
  for (long long i = 0; i < m; i++) {
    long long val = n - 2;
    long long ind = i;
    for (long long j = 19; j >= 0; j--) {
      if (1 << j <= val) {
        ind = dp[ind][j];
        val -= 1 << j;
        if (ind == oo) break;
      }
    }
    if (ind != oo)
      a[i] = a[ind];
    else
      a[i] = oo;
  }
  build(a);
  string ans;
  for (long long i = 0; i < k; i++) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    long long mn = get(1, 0, N - 1, l, r);
    if (mn <= r) {
      ans += '1';
    } else {
      ans += '0';
    }
  }
  cout << ans;
  return 0;
}
