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;
using lint = long long;
template <class T = int>
using V = vector<T>;
template <class T = int>
using VV = V<V<T> >;
struct SegmentTree {
  using T = int;
  static T op(const T& x, const T& y) { return min(x, y); }
  static constexpr T e() { return numeric_limits<T>::max(); }
  const int n;
  V<T> t;
  SegmentTree(int n) : n(n), t(2 * n, e()) {}
  T& operator[](int i) { return t[i + n]; }
  void build() {
    for (int i = n - 1; i; --i) t[i] = op(t[2 * i], t[2 * i + 1]);
  }
  T acc(int l, int r) const {
    T resl = e(), resr = e();
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) resl = op(resl, t[l++]);
      if (r & 1) resr = op(t[--r], resr);
    }
    return op(resl, resr);
  }
  void set(int i, const T& x) {
    t[i += n] = x;
    while (i >>= 1) t[i] = op(t[2 * i], t[2 * i + 1]);
  }
};
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  V<> ip(n);
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p, --p;
    ip[p] = i;
  }
  V<> a(m);
  for (auto&& e : a) cin >> e, --e, e = ip[e];
  V<> mn(n, m);
  VV<> par(20, V<>(m, m));
  for (int i = m - 1; i >= 0; --i) {
    par[0][i] = mn[(a[i] + 1) % n];
    mn[a[i]] = i;
  }
  for (int k = 1; k < (int)par.size(); ++k) {
    for (int i = 0; i < m; ++i) {
      if (par[k - 1][i] == m) continue;
      par[k][i] = par[k - 1][par[k - 1][i]];
    }
  }
  auto get_par = [&](int i, int h) -> int {
    for (int k = 0; h > 0; h >>= 1, ++k) {
      if (i == m) break;
      if (h & 1) i = par[k][i];
    }
    return i;
  };
  SegmentTree st(m);
  for (int i = 0; i < m; ++i) {
    st[i] = get_par(i, n - 1);
  }
  st.build();
  while (q--) {
    int l, r;
    cin >> l >> r, --l;
    cout << (st.acc(l, r) < r);
  }
  cout << '\n';
}
