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 n, m, q;
const int maxN = 2 * (int)1e5 + 100;
int p[maxN];
int a[maxN];
vector<int> all[maxN];
int nxt[maxN];
int l[maxN], r[maxN];
int nxt_pos[maxN];
int ansR[maxN];
int ansL[maxN];
bool ok[maxN];
int prv[maxN];
void solve1() {
  int best = 1;
  for (int i = 1; i <= n; i++) {
    if (all[i].size() < all[best].size()) {
      best = i;
    }
  }
  for (auto t : all[best]) {
    int x = best;
    ansR[t] = ansL[t] = 1;
    int need = nxt[best];
    for (int j = t + 1; j <= m; j++) {
      ansR[j] = ansR[j - 1];
      if (a[j] == need) {
        need = nxt[a[j]];
        ansR[j]++;
      }
    }
    need = prv[best];
    for (int j = t - 1; j >= 1; j--) {
      ansL[j] = ansL[j + 1];
      if (a[j] == need) {
        need = prv[a[j]];
        ansL[j]++;
      }
    }
    for (int i = 1; i <= q; i++) {
      if (ok[i]) continue;
      if (t < l[i] || t > r[i]) continue;
      if (ansR[r[i]] + ansL[l[i]] - 1 >= n) ok[i] = true;
    }
  }
  for (int i = 1; i <= q; i++) cout << ok[i];
  cout << '\n';
}
int ans[maxN];
int t[4 * maxN];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = ans[tl];
    return;
  }
  int tm = (tl + tr) / 2;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
  t[v] = min(t[v + v], t[v + v + 1]);
}
int get(int v, int tl, int tr, int l, int r) {
  assert(tl <= tr);
  if (tl == l && tr == r) return t[v];
  int tm = (tl + tr) / 2;
  if (r <= tm)
    return get(v + v, tl, tm, l, r);
  else if (l > tm)
    return get(v + v + 1, tm + 1, tr, l, r);
  else
    return min(get(v + v, tl, tm, l, tm),
               get(v + v + 1, tm + 1, tr, tm + 1, r));
}
vector<int> link[maxN];
vector<pair<int, int> > byR[maxN];
const int LOG = 20;
int up[maxN][LOG];
void solve2() {
  for (int i = m; i >= 1; i--) {
    up[i][0] = nxt_pos[i];
    for (int j = 0; j + 1 < LOG; j++) {
      up[i][j + 1] = up[up[i][j]][j];
    }
    int cur = i;
    int t = n - 1;
    for (int j = LOG - 1; j >= 0; j--) {
      if (t >= (1 << j)) {
        t -= (1 << j);
        cur = up[cur][j];
      }
    }
    if (cur == 0)
      ans[i] = (int)1e9;
    else
      ans[i] = cur;
  }
  build(1, 1, m);
  for (int i = 1; i <= q; i++) {
    if (get(1, 1, m, l[i], r[i]) <= r[i])
      cout << 1;
    else
      cout << 0;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; i++) {
    nxt[p[i]] = p[i % n + 1];
    prv[p[i]] = p[(i + 2 * n - 2) % n + 1];
  }
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    all[a[i]].push_back(i);
  }
  for (int i = 1; i <= m; i++) {
    nxt_pos[i] = 0;
    if (all[nxt[a[i]]].empty() || all[nxt[a[i]]].back() <= i) continue;
    int ind = upper_bound(all[nxt[a[i]]].begin(), all[nxt[a[i]]].end(), i) -
              all[nxt[a[i]]].begin();
    nxt_pos[i] = all[nxt[a[i]]][ind];
    link[nxt_pos[i]].push_back(i);
  }
  for (int i = 1; i <= q; i++) {
    cin >> l[i] >> r[i];
    byR[r[i]].emplace_back(l[i], i);
  }
  solve2();
  return 0;
}
