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 int INF = 9223372036854775807;
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
int max3(int a, int b, int c) { return max(c, max(a, b)); }
int min3(int a, int b, int c) { return min(a, min(b, c)); }
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x) { return power(x, 1000000007 - 2); }
vector<vector<long long int> > dp;
vector<long long int> ans;
vector<long long int> tree;
void build(int node, int start, int end) {
  if (start == end) {
    tree[node] = ans[start];
  } else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = min(tree[2 * node], tree[2 * node + 1]);
  }
}
long long int query(int node, int start, int end, int l, int r) {
  if (r < start or end < l) {
    return INF;
  }
  if (l <= start and end <= r) {
    return tree[node];
  }
  int mid = (start + end) / 2;
  long long int p1 = query(2 * node, start, mid, l, r);
  long long int p2 = query(2 * node + 1, mid + 1, end, l, r);
  return min(p1, p2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, m, q;
  cin >> n >> m >> q;
  map<int, int> g;
  tree.resize(4 * m + 50);
  vector<int> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    g[p[i]] = i;
  }
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  dp.resize(m);
  for (int i = 0; i < m; i++) {
    dp[i].resize(21, INF);
  }
  vector<vector<int> > pos;
  pos.resize(n + 1);
  for (int i = 0; i < m; i++) {
    pos[a[i]].push_back(i);
  }
  vector<long long int> par(m);
  for (int i = 0; i < m; i++) {
    int ind = g[a[i]];
    ind = (ind + 1) % n;
    int x = p[ind];
    auto it = upper_bound(pos[x].begin(), pos[x].end(), i);
    if (it != pos[x].end()) {
      par[i] = *it;
    } else {
      par[i] = INF;
    }
  }
  for (int j = m - 1; j >= 0; j--) {
    dp[j][0] = par[j];
    for (int i = 1; i <= 20; i++) {
      if (j + (1 << i) - 1 > m - 1) {
        dp[j][i] = INF;
      } else {
        if (dp[j][i - 1] == INF) {
          dp[j][i] = INF;
        } else {
          dp[j][i] = dp[dp[j][i - 1]][i - 1];
        }
      }
    }
  }
  ans.resize(m);
  for (int i = 0; i < m; i++) {
    long long int cur = i;
    long long int next = n - 1;
    if (next == 0) {
      ans[i] = i;
      continue;
    }
    for (int j = 20; j >= 0 && n > 0; j--) {
      if ((1 << j) < next && cur != INF) {
        cur = dp[cur][j];
        next -= 1 << j;
      }
    }
    if (cur != INF) {
      ans[i] = dp[cur][0];
    } else {
      ans[i] = INF;
    }
  }
  build(1, 0, m - 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (query(1, 0, m - 1, l - 1, r - 1) <= r - 1) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
}
