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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
using namespace std;
long long spf[301];
long long fac[301];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 301; i++) spf[i] = i;
  for (long long i = 4; i < 301; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 301; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 301; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r > n) return 0;
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
long long seg[4000009] = {0};
void updateseg(long long v, long long l, long long r, long long pos,
               long long val) {
  if (pos > r || pos < l) return;
  if (l == r) {
    seg[v] = max(val, seg[v]);
    return;
  }
  long long mid = (l + r) / 2;
  if (pos <= mid)
    updateseg(v * 2, l, mid, pos, val);
  else
    updateseg(v * 2 + 1, mid + 1, r, pos, val);
  seg[v] = max(seg[v * 2], seg[v * 2 + 1]);
}
long long queryseg(long long v, long long l, long long r, long long posl,
                   long long posr) {
  if (l == posl && r == posr)
    return seg[v];
  else if (posl > posr)
    return 0;
  long long mid = (l + r) / 2;
  return max(queryseg(v * 2, l, mid, posl, min(posr, mid)),
             queryseg(v * 2 + 1, mid + 1, r, max(posl, mid + 1), posr));
}
multiset<pair<long long, long long> > s[200001];
map<long long, long long> m;
long long dp[200005];
vector<pair<long long, long long> > v[200001];
long long ans[200001] = {0};
void solve() {
  long long n, ma, q;
  cin >> n >> ma >> q;
  long long a[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    m[a[i]] = i;
  }
  long long b[ma + 1];
  for (long long i = 1; i <= ma; i++) cin >> b[i];
  for (long long i = 1; i <= 200000; i++) dp[i] = 1e18;
  for (long long i = ma; i >= 1; i--) {
    long long ind = m[b[i]];
    if (ind == n)
      dp[n] = i;
    else
      dp[ind] = min(dp[ind], dp[ind + 1]);
    s[ind].insert(make_pair(dp[ind], i));
  }
  memset(dp, 0, sizeof(dp));
  for (long long i = 1; i <= q; i++) {
    long long l, r;
    cin >> l >> r;
    v[r].push_back(make_pair(l, i));
  }
  for (long long i = 1; i <= ma; i++) {
    long long ind = m[b[i]];
    if (ind == 1)
      dp[ind] = i;
    else
      dp[ind] = max(dp[ind], dp[ind - 1]);
    if (ind == n) {
      updateseg(1, 1, ma, dp[ind], 1);
    } else {
      auto it = s[ind + 1].lower_bound(make_pair(dp[ind], 0));
      if (it != s[ind + 1].begin() && s[ind + 1].size()) {
        auto it2 = it;
        it2--;
        updateseg(1, 1, ma, it2->second, dp[ind] - it2->first);
      }
    }
    for (auto it : v[i]) {
      if (queryseg(1, 1, ma, it.first, i))
        ans[it.second] = 1;
      else
        ans[it.second] = 0;
    }
  }
  for (long long i = 1; i <= q; i++) cout << ans[i];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
}
