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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
template <typename T, typename OP>
struct seg_tree {
  OP op;
  T *t;
  int n;
  int L, R;
  vector<T> &v;
  T build(int l, int r, int i) {
    if (l == r) return t[i] = op(v[l]);
    int m = (l + r) / 2;
    T lb = build(l, m, 2 * i + 1);
    T rb = build(m + 1, r, 2 * i + 2);
    return t[i] = op(lb, rb);
  }
  T query(int l, int r, int i) {
    if (L <= l && r <= R) return t[i];
    int m = (l + r) / 2;
    if (m >= R) return query(l, m, 2 * i + 1);
    if (m < L) return query(m + 1, r, 2 * i + 2);
    return op(query(l, m, 2 * i + 1), query(m + 1, r, 2 * i + 2));
  }
  T query(int L_, int R_) {
    L = L_;
    R = R_;
    return query(0, n - 1, 0);
  }
  T update(int l, int r, int i, int k) {
    if (l == r) return t[i] = op(v[k]);
    int m = (l + r) / 2;
    if (m >= k) return t[i] = op(t[2 * i + 2], update(l, m, 2 * i + 1, k));
    if (m < k) return t[i] = op(t[2 * i + 1], update(m + 1, r, 2 * i + 2, k));
    return t[i] =
               op(update(l, m, 2 * i + 1, k), update(m + 1, r, 2 * i + 2, k));
  }
  T update(int k) { return update(0, n - 1, 0, k); }
  seg_tree(vector<T> &v_) : v(v_) {
    n = v.size();
    t = new T[4 * n + 1];
    build(0, n - 1, 0);
  }
  ~seg_tree() { delete[] t; }
  seg_tree() {}
};
template <typename T>
struct max_el {
  T operator()(const T &l) { return l; }
  T operator()(const T &l, const T &r) { return max(l, r); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> p(n), a(m);
  for (int &i : p) cin >> i;
  for (int &i : a) cin >> i;
  vector<int> prev_p(n + 1);
  prev_p[p[0]] = p[n - 1];
  for (int i = 1; i < n; i++) prev_p[p[i]] = p[i - 1];
  vector<vector<int>> prev(m, vector<int>(20, -1));
  vector<int> last_occ(n + 1, -1);
  for (int i = 0; i < m; i++) {
    prev[i][0] = last_occ[prev_p[a[i]]];
    last_occ[a[i]] = i;
  }
  for (int i = 1; i < 20; i++)
    for (int j = 0; j < m; j++)
      if (prev[j][i - 1] != -1) prev[j][i] = prev[prev[j][i - 1]][i - 1];
  auto occ = [&](int p, int n) {
    int ans = p;
    for (int it = 19; it >= 0; it--) {
      if (n & (1 << it)) {
        ans = prev[ans][it];
      }
      if (ans == -1) break;
    }
    return ans;
  };
  vector<int> t(m + 1, 0);
  for (int i = 0; i < m; i++) t[i] = occ(i, n - 1);
  seg_tree<int, max_el<int>> st(t);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    int pos = st.query(l, r);
    if (pos >= l)
      cout << 1;
    else
      cout << 0;
  }
  cout << endl;
  exit(0);
}
