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;
vector<vector<long long int> > matMul(vector<vector<long long int> > A,
                                      vector<vector<long long int> > B,
                                      long long int m) {
  int r1 = A.size(), r2 = B.size(), c1 = A[0].size(), c2 = B[0].size();
  vector<vector<long long int> > ans(r1);
  for (int i = 0; i < r1; i++)
    for (int j = 0; j < c2; j++) {
      long long int mul = 0;
      for (int k = 0; k < c1; k++) mul = (mul + A[i][k] * B[k][j]) % m;
      ans[i].push_back(mul);
    }
  return ans;
}
vector<vector<long long int> > matPow(vector<vector<long long int> > first,
                                      long long int n, long long int m) {
  int r = first.size(), c = first[0].size();
  vector<vector<long long int> > result(r);
  for (int i = int(0); i < int(r); i++)
    for (int j = int(0); j < int(c); j++) {
      if (i == j)
        result[i].push_back(1);
      else
        result[i].push_back(0);
    }
  while (n > 0) {
    if (n % 2 == 1) result = matMul(result, first, m);
    first = matMul(first, first, m);
    n = n / 2;
  }
  return result;
}
long long int mod = 1e9 + 7;
long long int pow1(long long int first, long long int n, long long int m) {
  long long int result = 1;
  while (n > 0) {
    first %= m;
    if (n % 2 == 1) result = (result * first) % m;
    first = (first * first) % m;
    n = n / 2;
  }
  return result;
}
long long int pow2(long long int first, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * first);
    first = (first * first);
    n = n / 2;
  }
  return result;
}
long long int modInverse(long long int A, long long int M) {
  return pow1(A, M - 2, M);
}
long long int f[1000001];
void fact() {
  f[0] = f[1] = 1;
  for (int i = 2; i <= 1e6; i++) f[i] = (f[i - 1] * i) % mod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
const int size = 1e6 + 1;
int n, m, q, a[size], b[size], my[size], par[200001][25], len[size];
vector<int> pos[size];
int fun(int id) {
  int z = n - 1;
  for (int i = 25; i >= 0; i--) {
    if ((1 << i) & z) {
      id = par[id][i];
    }
    if (id == m) break;
  }
  return id;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> q;
  for (int i = int(0); i < int(n); i++) {
    cin >> a[i];
    my[a[i]] = i;
  }
  for (int i = int(0); i < int(m); i++) {
    cin >> b[i];
    pos[b[i]].push_back(i);
  }
  for (int i = int(0); i < int(m); i++) {
    int next = my[b[i]] == n - 1 ? a[0] : a[my[b[i]] + 1];
    int id =
        upper_bound(pos[next].begin(), pos[next].end(), i) - pos[next].begin();
    if (id == pos[next].size())
      par[i][0] = m;
    else
      par[i][0] = pos[next][id];
  }
  for (int i = 1; (1 << (i - 1)) <= m; i++) {
    for (int j = 0; j < m; j++) {
      if (par[j][i - 1] == m)
        par[j][i] = m;
      else
        par[j][i] = par[par[j][i - 1]][i - 1];
    }
  }
  for (int i = int(0); i < int(m); i++) {
    len[i] = fun(i);
  }
  int mi[m];
  mi[m - 1] = len[m - 1];
  for (int i = m - 2; i >= 0; i--) {
    mi[i] = min(mi[i + 1], len[i]);
  }
  string ans = "";
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    if (mi[l] <= r)
      ans += "1";
    else
      ans += "0";
  }
  cout << ans;
  return 0;
}
