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 = 100008;
int n, q, l, r, Last[maxn], Next[maxn], Head[maxn], cnt[maxn], d[maxn], a[maxn],
    fl[maxn], fr[maxn], ccol, cq, sqrtn, ans[maxn];
bool Good[maxn];
struct Query {
  int l, r, o;
  bool operator<(const Query u) const {
    return l / sqrtn == u.l / sqrtn ? r < u.r : l < u.l;
  }
} Q[maxn];
inline void Modify(int v, int k) {
  if (k)
    if (!Good[v]) cq++, Good[v] = true;
  if (!k)
    if (Good[v]) cq--, Good[v] = false;
}
inline void Insert(int p, int k) {
  int v = a[p], _d;
  cnt[v]++;
  if (cnt[v] == 1)
    ccol++, Modify(v, 1);
  else if (k ? fl[p] >= l : fr[p] <= r)
    Modify(v, 0);
}
inline void Delete(int p, int k) {
  int v = a[p], _d;
  cnt[v]--;
  if (cnt[v] == 0)
    ccol--, Modify(v, 0);
  else if (k ? fl[Last[p]] < l : fr[Next[p]] > r)
    Modify(v, 1);
}
int main() {
  scanf("%d", &n);
  sqrtn = 300;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  fill(Head, Head + maxn, 0);
  for (int i = 1; i <= n; i++) {
    Last[i] = Head[a[i]], Head[a[i]] = i;
    if (Last[i] - Last[Last[i]] == i - Last[i] || !Last[Last[i]])
      fl[i] = fl[Last[i]];
    else
      fl[i] = Last[Last[i]];
  }
  fill(Head, Head + maxn, n + 1);
  Next[n + 1] = n + 1;
  fr[n + 1] = n + 1;
  for (int i = n; i; i--) {
    Next[i] = Head[a[i]], Head[a[i]] = i;
    if (Next[Next[i]] - Next[i] == Next[i] - i || Next[Next[i]] == n + 1)
      fr[i] = fr[Next[i]];
    else
      fr[i] = Next[Next[i]];
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) scanf("%d%d", &Q[i].l, &Q[i].r), Q[i].o = i;
  sort(Q + 1, Q + q + 1);
  l = 1;
  for (int i = 1; i <= q; i++) {
    int _l = Q[i].l, _r = Q[i].r;
    while (r < _r) Insert(++r, 1);
    while (l > _l) Insert(--l, 0);
    while (r > _r) Delete(r--, 1);
    while (l < _l) Delete(l++, 0);
    ans[Q[i].o] = ccol + (int)(!cq);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
