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 long long N = 2e5 + 10;
long long P[N];
long long A[N];
vector<long long> pos[N];
vector<long long> adj[N];
long long anc[N][20];
long long T[N];
long long L[N];
long long pre[N];
long long ST[N][20];
void dfs(long long u) {
  for (long long v : adj[u]) {
    if (L[v] == -1) {
      T[v] = u;
      L[v] = L[u] + 1;
      dfs(v);
    }
  }
  return;
}
void build(long long n) {
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; (1ll << j) <= n; j++) {
      anc[i][j] = -1;
    }
  }
  for (long long i = 0; i <= n; i++) {
    anc[i][0] = T[i];
  }
  for (long long j = 1; (1ll << j) <= n; j++) {
    for (long long i = 0; i <= n; i++) {
      if (anc[i][j - 1] != -1) {
        anc[i][j] = anc[anc[i][j - 1]][j - 1];
      }
    }
  }
  return;
}
void une(long long a, long long b) {
  for (long long x : pos[a]) {
    long long p = lower_bound(pos[b].begin(), pos[b].end(), x) - pos[b].begin();
    p--;
    if (p >= 0) {
      adj[pos[b][p]].push_back(x);
    }
  }
  return;
}
void precalculo(long long n) {
  for (long long i = 0; i <= n; i++) {
    ST[i][0] = pre[i];
  }
  for (long long j = 1; (1 << j) <= n; j++) {
    for (long long i = 0; i + (1 << j) - 1 <= n; i++) {
      ST[i][j] = max(ST[i][j - 1], ST[i + (1 << (j - 1))][j - 1]);
    }
  }
  return;
}
long long query(long long l, long long r) {
  long long T = r - l + 1;
  long long lg = 63 - (__builtin_clzll(T));
  return max(ST[l][lg], ST[r - (1 << lg) + 1][lg]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long n, m, q;
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> P[i];
  }
  P[0] = P[n];
  for (long long i = 1; i <= m; i++) {
    cin >> A[i];
    pos[A[i]].push_back(i);
  }
  for (long long i = n; i > 0; i--) {
    une(P[i], P[i - 1]);
  }
  for (long long i = 0; i <= m; i++) {
    T[i] = -1;
    L[i] = -1;
  }
  for (long long i = 0; i <= m; i++) {
    if (L[i] == -1) {
      L[i] = 0;
      dfs(i);
    }
  }
  build(m);
  n--;
  for (long long i = 1; i <= m; i++) {
    long long pp = i;
    for (long long j = 0; j < 20; j++) {
      if (((n >> j) & 1) == 1) {
        pp = anc[pp][j];
      }
      if (pp == -1) {
        break;
      }
    }
    pre[i] = pp;
  }
  pre[0] = -1;
  precalculo(m);
  long long x, y;
  for (long long i = 0; i < q; i++) {
    cin >> x >> y;
    if (query(x, y) >= x) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  cout << endl;
  return 0;
}
