Cosider a sequence, consisting of n integers: a1, a2, ..., an. Jeff can perform the following operation on sequence a:

  * take three integers v, t, k (1 ≤ v, t ≤ n; 0 ≤ k; v + tk ≤ n), such that av = av + t, av + t = av + 2t, ..., av + t(k - 1) = av + tk; 
  * remove elements av, av + t, ..., av + t·k from the sequence a, the remaining elements should be reindexed a1, a2, ..., an - k - 1. 
  * permute in some order the remaining elements of sequence a. 



A beauty of a sequence a is the minimum number of operations that is needed to delete all elements from sequence a.

Jeff's written down a sequence of m integers b1, b2, ..., bm. Now he wants to ask q questions. Each question can be described with two integers li, ri. The answer to the question is the beauty of sequence bli, bli + 1, ..., bri. You are given the sequence b and all questions. Help Jeff, answer all his questions.

Input

The first line contains integer m (1 ≤ m ≤ 105). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 105). 

The third line contains integer q (1 ≤ q ≤ 105) — the number of questions. The next q lines contain pairs of integers, i-th of them contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ m) — the description of i-th question.

Output

In q lines print the answers to Jeff's queries. Print the answers according to the order of questions in input.

Examples

Input

5
2 2 1 1 2
5
1 5
1 1
2 2
1 3
2 3


Output

2
1
1
2
2


Input

10
2 1 3 3 3 3 1 3 1 1
10
4 8
2 10
1 10
4 4
1 3
2 4
6 7
1 9
2 5
1 1


Output

2
3
3
1
3
2
2
3
2
1
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int sqrtn = 505;
int l = 1, r, sum, cant;
int n, m, p, s[maxn], w[maxn], cnt[maxn], last[maxn], pre[maxn], nxt[maxn],
    qwq[maxn][sqrtn], num[maxn];
vector<int> v;
struct node {
  int l, r, id;
} a[maxn];
inline bool cmp(node a, node b) {
  return a.l / p == b.l / p ? ((a.l / p) & 1 ? a.r < b.r : a.r > b.r)
                            : a.l < b.l;
}
inline void add(int x) {
  cnt[w[x]]++;
  sum += cnt[w[x]] == 1;
}
inline void add_right(int x) {
  add(x);
  if (cnt[w[x]] != 1 && pre[x]) {
    qwq[w[x]][pre[x]]++;
    if (qwq[w[x]][pre[x]] == 1) {
      num[w[x]]++;
      if (num[w[x]] == 2) {
        cant++;
      }
    }
  }
}
void add_left(int x) {
  add(x);
  if (cnt[w[x]] != 1 && nxt[x]) {
    qwq[w[x]][nxt[x]]++;
    if (qwq[w[x]][nxt[x]] == 1) {
      num[w[x]]++;
      if (num[w[x]] == 2) {
        cant++;
      }
    }
  }
}
inline void del(int x) {
  cnt[w[x]]--;
  sum -= cnt[w[x]] == 0;
}
inline void del_right(int x) {
  del(x);
  if (qwq[w[x]][pre[x]] && pre[x]) {
    qwq[w[x]][pre[x]]--;
    if (qwq[w[x]][pre[x]] == 0) {
      num[w[x]]--;
      if (num[w[x]] == 1) {
        cant--;
      }
    }
  }
}
inline void del_left(int x) {
  del(x);
  if (qwq[w[x]][nxt[x]] && nxt[x]) {
    qwq[w[x]][nxt[x]]--;
    if (qwq[w[x]][nxt[x]] == 0) {
      num[w[x]]--;
      if (num[w[x]] == 1) {
        cant--;
      }
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    if (last[w[i]]) {
      pre[i] = i - last[w[i]];
      nxt[last[w[i]]] = i - last[w[i]];
      v.push_back(pre[i]);
      v.push_back(nxt[last[w[i]]]);
    }
    last[w[i]] = i;
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; i++) {
  }
  p = sqrt(n);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i].l >> a[i].r;
    a[i].id = i;
  }
  sort(a + 1, a + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    while (l > a[i].l) {
      add_left(--l);
    }
    while (r < a[i].r) {
      add_right(++r);
    }
    while (l < a[i].l) {
      del_left(l++);
    }
    while (r > a[i].r) {
      del_right(r--);
    }
    s[a[i].id] = sum + (cant == sum);
  }
  for (int i = 1; i <= m; i++) {
    cout << s[i] << "\n";
  }
}
