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 int maxn = (int)1e5;
int p[maxn * 2 + 1], a[maxn * 2], n, m;
int pos[maxn * 2 + 1], Next[2 * maxn];
int last[maxn * 2], last_meet[maxn * 2 + 1];
const int ts = 1 << 18;
int st[2 * ts];
int nxt(int i) {
  if (i == n - 1) return 0;
  return i + 1;
}
vector<int> g[maxn * 2];
int us[maxn * 2], Stack[maxn * 2], sstack_size = 0;
void dfs(int i) {
  Stack[sstack_size++] = i;
  if (sstack_size >= n) last[i] = Stack[sstack_size - n];
  us[i] = 1;
  for (auto u : g[i]) dfs(u);
  --sstack_size;
}
void not_one() {
  for (int i = 0; i < n; ++i) pos[p[i]] = i;
  fill(last_meet, last_meet + n + 1, m);
  for (int i = m - 1; i > -1; --i) {
    Next[i] = last_meet[p[nxt(pos[a[i]])]];
    last_meet[a[i]] = i;
  }
  for (int i = 0; i < m; ++i) {
    if (Next[i] != m) g[Next[i]].push_back(i);
  }
  for (int i = m - 1; i > -1; --i)
    if (!us[i]) dfs(i);
}
void one() {
  for (int i = 0; i < m; ++i)
    if (a[i] == p[0]) last[i] = i;
}
void build() {
  fill(last, last + m, m);
  if (n != 1)
    not_one();
  else
    one();
  for (int i = 0; i < m; ++i) st[i + ts] = last[i] + 1;
  for (int i = ts - 1; i; --i) st[i] = min(st[2 * i], st[2 * i + 1]);
}
char query(int l, int r) {
  if (r - l + 1 < n) return '0';
  int permis = r;
  l += ts - 1;
  r += ts - 1;
  int mn = m + 1;
  while (l <= r) {
    if (l & 1) mn = min(mn, st[l++]);
    if (!(r & 1)) mn = min(mn, st[r--]);
    l /= 2;
    r /= 2;
  }
  return '0' + (mn <= permis);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> m >> q;
  for (int i = 0; i < n; ++i) cin >> p[i];
  for (int i = 0; i < m; ++i) cin >> a[i];
  build();
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << query(l, r);
  }
  return 0;
}
