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 GCC optimize("-O3", "unroll-loops")
using namespace std;
template <typename T1, typename T2>
inline void chmin(T1& a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1& a, T2 b) {
  if (a < b) a = b;
}
const long long MOD = 1e9 + 7;
const long long MAX = 200010;
const double pi = acos(-1);
const long long MAX_N = 1 << 21;
const long long MAX_V = 200010;
long long n, dat[2 * MAX_N - 1];
long long id[MAX_V];
void init(long long n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (long long i = 0; i < 2 * n - 1; i++) dat[i] = INT_MAX;
}
void update(long long k, long long a) {
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
  }
}
long long query(long long a, long long b, long long k, long long l,
                long long r) {
  if (r <= a || b <= l) return INT_MAX;
  if (a <= l && r <= b)
    return dat[k];
  else {
    long long vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    long long vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
}
set<long long> st[200020];
long long p[200020], A[200020];
long long nxt[200020][30];
long long next_number[200020];
void solve() {
  long long N, M, Q;
  cin >> N >> M >> Q;
  for (long long i = 0; i < N; i++) cin >> p[i];
  for (long long i = 0; i < N; i++) {
    next_number[p[i]] = p[(i + 1) % N];
  }
  for (long long i = 0; i < M; i++) {
    cin >> A[i];
    st[A[i]].insert(i);
  }
  for (long long i = 0; i < M; i++) {
    long long num = next_number[A[i]];
    auto itr = st[num].upper_bound(i);
    if (itr == st[num].end()) {
      nxt[i][0] = M;
    } else
      nxt[i][0] = *itr;
  }
  init(M + 1);
  for (long long d = 0; d < 25; d++) {
    nxt[M][d] = M;
    for (long long i = 0; i < M; i++) {
      long long tmp = nxt[i][d];
      nxt[i][d + 1] = nxt[tmp][d];
    }
  }
  vector<long long> res(M);
  for (long long i = 0; i < M; i++) {
    long long idx = i;
    long long rem = N - 1;
    for (long long d = 24; d >= 0; d--) {
      if ((1 << d) <= rem) {
        rem -= (1 << d);
        idx = nxt[idx][d];
      }
    }
    res[i] = idx;
    update(i, idx);
  }
  string ans;
  for (long long i = 0; i < Q; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    long long res = query(l, r, 0, 0, n);
    if (res < r) {
      ans += '1';
    } else
      ans += '0';
  }
  cout << ans << '\n';
  ;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}
