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;
int perm[1000005];
int dok[1000005];
vector<int> gdzie[1000005];
int tab[1000005];
vector<int> pol[1000005];
int skoki[25][200005];
bool odw[200005];
int pot = 1 << 20;
int drzewo[4000005];
void ins(int a, int b) {
  int pom = a + pot;
  while (pom != 0) {
    drzewo[pom] = min(drzewo[pom], b);
    pom /= 2;
  }
}
int check(int a, int b) {
  int war = 1e9, l = a + pot, r = b + pot;
  war = drzewo[l];
  if (l != r) war = min(war, drzewo[r]);
  while (l / 2 != r / 2) {
    if (l % 2 == 0) war = min(war, drzewo[l + 1]);
    if (r % 2 == 1) war = min(war, drzewo[r - 1]);
    l /= 2;
    r /= 2;
  }
  return war;
}
void dfs(int a) {
  for (auto x : pol[a]) {
    if (!odw[x]) {
      skoki[0][x] = a;
      odw[x] = true;
      dfs(x);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int a, b, t;
  cin >> a >> b >> t;
  for (int x = 0; x < a; x++) cin >> perm[x];
  for (int x = 0; x < a; x++) dok[perm[x]] = perm[(x + 1) % a];
  for (int x = 0; x < b; x++) {
    cin >> tab[x];
    gdzie[tab[x]].push_back(x);
  }
  for (int x = 1; x <= a; x++) gdzie[x].push_back(b);
  for (int x = 0; x < b; x++) {
    pol[x].push_back(
        *lower_bound(gdzie[dok[tab[x]]].begin(), gdzie[dok[tab[x]]].end(), x));
    pol[*lower_bound(gdzie[dok[tab[x]]].begin(), gdzie[dok[tab[x]]].end(), x)]
        .push_back(x);
  }
  odw[b] = true;
  dfs(b);
  skoki[0][b] = b;
  for (int i = 1; i < 21; i++)
    for (int j = 0; j <= b; j++) skoki[i][j] = skoki[i - 1][skoki[i - 1][j]];
  for (int x = 0; x <= 4000000; x++) drzewo[x] = 1e9;
  int pom = a - 1, pom1, pom2, pom3;
  for (int x = 0; x < b; x++) {
    pom1 = x;
    pom2 = 1;
    pom3 = 0;
    while (pom2 < 2 * 1e6) {
      if ((pom2 & pom) != 0) pom1 = skoki[pom3][pom1];
      pom3++;
      pom2 *= 2;
    }
    ins(x, pom1);
  }
  int c, d;
  while (t--) {
    cin >> c >> d;
    if (check(c - 1, d - 1) <= d - 1)
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
