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;
struct Node {
  int l, r, whe;
} s[100100];
struct Point {
  int v, whe;
} a[100100];
int m, q, sqr, R[100100], L[100100];
int cnt[100100], ncol[100100], l, r, ans[100100], tot, tot2;
bool cmp(Node x, Node y) {
  return x.l / sqr < y.l / sqr || (x.l / sqr == y.l / sqr && x.r < y.r);
}
bool cmp2(Point x, Point y) {
  return x.v < y.v || (x.v == y.v && x.whe < y.whe);
}
bool cmp3(Point x, Point y) { return x.whe < y.whe; }
void prep() {
  sort(a + 1, a + m + 1, cmp2);
  for (int i = 1; i <= m; i++) {
    if (a[i].v != a[i - 1].v || a[i].v != a[i - 2].v) {
      L[a[i].whe] = -1;
      R[a[i - 2].whe] = -1;
      continue;
    }
    if (a[i].whe - a[i - 1].whe != a[i - 1].whe - a[i - 2].whe)
      L[a[i].whe] = a[i - 2].whe, R[a[i - 2].whe] = a[i].whe;
    else
      L[a[i].whe] = -1, R[a[i - 2].whe] = -1;
  }
  R[a[m - 1].whe] = -1;
  R[a[m].whe] = -1;
  sort(a + 1, a + m + 1, cmp3);
}
void into(int x) {
  if (!ncol[a[x].v]) tot++;
  ncol[a[x].v]++;
  if (L[x] != -1 && L[x] >= l) {
    if (!cnt[a[x].v]) tot2++;
    cnt[a[x].v]++;
  }
  if (R[x] != -1 && R[x] <= r) {
    if (!cnt[a[x].v]) tot2++;
    cnt[a[x].v]++;
  }
}
void outit(int x) {
  ncol[a[x].v]--;
  if (!ncol[a[x].v]) tot--;
  if (L[x] != -1 && L[x] >= l) {
    cnt[a[x].v]--;
    if (!cnt[a[x].v]) tot2--;
  }
  if (R[x] != -1 && R[x] <= r) {
    cnt[a[x].v]--;
    if (!cnt[a[x].v]) tot2--;
  }
}
void work() {
  l = s[1].l, r = s[1].l - 1;
  for (int i = 1; i <= q; i++) {
    while (l > s[i].l) l--, into(l);
    while (l < s[i].l) outit(l), l++;
    while (r > s[i].r) outit(r), r--;
    while (r < s[i].r) r++, into(r);
    if (tot == 1)
      ans[s[i].whe] = 0;
    else
      ans[s[i].whe] = tot - (tot != tot2);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i] + 1);
}
int main() {
  cin >> m;
  for (int i = 1; i <= m; i++) scanf("%d", &a[i].v), a[i].whe = i;
  cin >> q;
  for (int i = 1; i <= q; i++) scanf("%d %d", &s[i].l, &s[i].r), s[i].whe = i;
  sqr = sqrt(m);
  sort(s + 1, s + q + 1, cmp);
  prep();
  work();
}
