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;
unsigned long long MSB(unsigned long long x) {
  x |= x >> 0x01;
  x |= x >> 0x02;
  x |= x >> 0x04;
  x |= x >> 0x08;
  x |= x >> 0x10;
  x |= x >> 0x20;
  return (x >> 1) + 1;
}
template <typename T>
struct rtree {
  vector<T> v;
  T z;
  rtree(unsigned long long n, T x = T()) : v(MSB(n - 1) << 2, x), z(x) {}
  void set(unsigned long long i, T x) {
    for (i += v.size() / 2; i; i /= 2) {
      v[i] = x;
      x = min(v[i & ~1], v[i | 1]);
    }
  }
  T& raw(unsigned long long i) { return v[i + v.size() / 2]; }
  void update() {
    for (unsigned long long i = v.size() / 2 - 1; i; --i) {
      v[i] = min(v[2 * i], v[2 * i + 1]);
    }
  }
  T get0(unsigned long long b, unsigned long long s, unsigned long long e,
         unsigned long long i, unsigned long long j) {
    if (s == i && e == j) return v[b];
    unsigned long long m = (s + e) / 2;
    if (j <= m) return get0(2 * b, s, m, i, j);
    if (i >= m) return get0(2 * b + 1, m, e, i, j);
    return min(get0(2 * b, s, m, i, m), get0(2 * b + 1, m, e, m, j));
  }
  T get(unsigned long long i, unsigned long long j) {
    if (i >= j) return z;
    return get0(1, 0, v.size() / 2, i, j);
  }
};
struct tree {
  vector<int> parent, level;
  vector<vector<int>> ancestor;
  tree(int n)
      : parent(n + 1, -1),
        level(n + 1, -1),
        ancestor(n + 1, vector<int>(30, -1)) {
    level[0] = 0;
  }
  void insert(int i, int par) {
    parent[i] = par;
    level[i] = level[par] + 1;
    ancestor[i][0] = par;
    for (int j = 1; (1 << j) < parent.size(); ++j) {
      if (ancestor[i][j - 1] != -1)
        ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];
    }
  }
  int lca(int u, int v) {
    if (level[u] < level[v]) swap(u, v);
    int dist = level[u] - level[v];
    while (dist > 0) {
      int rb = (int)log2(dist);
      u = ancestor[u][rb];
      dist -= (1 << rb);
    }
    if (u == v) return u;
    for (int j = log2(parent.size()); j >= 0; --j) {
      if (ancestor[u][j] != -1 && ancestor[u][j] != ancestor[v][j]) {
        u = ancestor[u][j];
        v = ancestor[v][j];
      }
    }
    return parent[u];
  }
  int anc(int u, int dep) {
    while (u > 0 && dep > 0) {
      int lg = log2(dep);
      u = ancestor[u][lg];
      dep -= (1 << lg);
    }
    return u;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> p(n + 1), next(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; ++i) {
    next[p[i]] = p[i % n + 1];
  }
  vector<int> a(m + 1);
  vector<set<int>> loc(n + 1);
  for (int i = 1; i <= m; ++i) {
    cin >> a[i];
    loc[a[i]].insert(i);
  }
  tree t(m + 1);
  for (int i = m; i >= 1; --i) {
    auto it = loc[next[a[i]]].lower_bound(i + 1);
    if (it != loc[next[a[i]]].end()) {
      t.insert(i, *it);
    } else {
      t.insert(i, 0);
    }
  }
  rtree<long long> rt(m + 1, 2e5 + 2);
  for (int i = 1; i <= m; ++i) {
    int a = t.anc(i, n - 1);
    if (a > 0) {
      rt.raw(i) = a;
    }
  }
  rt.update();
  while (q-- > 0) {
    int l, r;
    cin >> l >> r;
    int ed = rt.get(l, r + 1);
    if (ed <= r)
      cout << 1;
    else
      cout << 0;
  }
  cout << endl;
}
