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;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF2 = 8e18;
const long long INF = 2e18;
const long long MOD = 1e9 + 7;
const long long LOG = 22;
long long mn_r[MAXN], nxt[MAXN], P[MAXN], A[MAXN], n, m, qr;
set<long long> ind[MAXN];
long long table[MAXN][LOG];
long long walk(long long i, long long k) {
  long long j = 0;
  long long ans = i;
  while (k > 0) {
    if (k & 1) ans = table[ans][j];
    k /= 2;
    j++;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fill(mn_r, mn_r + MAXN, INF2);
  cin >> n >> m >> qr;
  if (n == 1) {
    while (qr--) cout << 1;
    cout << '\n';
    return 0;
  }
  for (int i = 0; i < n; i++) cin >> P[i];
  nxt[P[n - 1]] = P[0];
  for (int i = 1; i < n; i++) nxt[P[i - 1]] = P[i];
  for (int i = 0; i < m; i++) cin >> A[i], ind[A[i]].insert(i);
  for (int i = 0; i < m; i++) {
    auto it = ind[nxt[A[i]]].upper_bound(i);
    if (it == ind[nxt[A[i]]].end())
      table[i][0] = m;
    else
      table[i][0] = *it;
  }
  fill(table[m], table[m] + LOG, m);
  for (int i = m - 1; i >= 0; i--) {
    long long j = 1;
    while (j < LOG) table[i][j] = table[table[i][j - 1]][j - 1], j++;
  }
  mn_r[m - 1] = m;
  for (int i = m - 2; i >= 0; i--) mn_r[i] = min(mn_r[i + 1], walk(i, n - 1));
  while (qr--) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    if (mn_r[l] <= r)
      cout << 1;
    else
      cout << 0;
  }
  cout << '\n';
  return 0;
}
