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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = (((1 << 30) - 1) << 1) + 1;
const int nINF = 1 << 31;
const int N = 2e5 + 10;
int ts[N], te[N], lcat = 0;
vector<int> g[N], ba[N];
void lcadfs(int s, int p = 0, int d = 0) {
  ts[s] = lcat++;
  ba[s].push_back(p);
  for (int i = 1; d - (1 << i) >= 0; ++i)
    ba[s].push_back(ba[ba[s][i - 1]][i - 1]);
  for (int v : g[s])
    if (v != p) lcadfs(v, s, d + 1);
  te[s] = lcat++;
}
int n, m, q, p[N], a[N], ear[N], leaf[N], b[N];
set<pair<int, int> > segs;
stack<int> lastids[N];
int jump_forward(int k) {
  int num = n - 1;
  for (int i = (0); i <= (20); i++) {
    if ((1 << i) & num) {
      if (i < ba[k].size())
        k = ba[k][i];
      else
        return 0;
    }
  }
  return k;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = (1); i <= (n); i++) cin >> p[i];
  b[p[1]] = p[n];
  for (int i = (2); i <= (n); i++) b[p[i]] = p[i - 1];
  for (int i = (1); i <= (m); i++) cin >> a[i];
  for (int i = (1); i <= (m); i++) {
    int bef = b[a[i]];
    stack<int>& s = lastids[bef];
    while (!s.empty()) {
      g[i].push_back(s.top());
      leaf[s.top()] = 1;
      s.pop();
    }
    lastids[a[i]].push(i);
  }
  for (int i = (1); i <= (m); i++)
    if (!leaf[i]) lcadfs(i);
  for (int i = (1); i <= (m); i++) ear[i] = jump_forward(i);
  for (int i = (1); i <= (m); i++) {
    if (ear[i]) {
      while (segs.size() && segs.rbegin()->second >= ear[i])
        segs.erase(*segs.rbegin());
      segs.insert({i, ear[i]});
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto itr = segs.lower_bound({l, 0});
    cout << !(itr == segs.end() || itr->second > r);
  }
  cout << endl;
  return 0;
}
