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;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MOD = 1000000007;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
int mx[8] = {-1, 1, 0, 0, -1, -1, 1, 1};
int my[8] = {0, 0, -1, 1, -1, 1, -1, 1};
struct query {
  int l, r, id;
  query() {}
  query(int l, int r, int id) : l(l), r(r), id(id) {}
};
int dv = 316;
bool cf(query p, query q) {
  if (p.l / dv != q.l / dv)
    return p.l / dv < q.l / dv;
  else
    return p.r < q.r;
}
int nx[100005];
int pr[100005];
int ls[100005];
int co[100005];
int x[100005];
int ans[100005];
query q[100005];
int on = 0;
map<int, int> dif[100005];
int nyala[100005];
int can = 0;
int single = 0;
void add(int k) {
  k = x[k];
  if (co[k] == 0) on++;
  if (co[k] == 1) single--;
  co[k]++;
  if (co[k] == 1) single++;
}
void rem(int k) {
  k = x[k];
  if (co[k] == 1) on--;
  if (co[k] == 1) single--;
  co[k]--;
  if (co[k] == 1) single++;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int(a) = (1); (a) <= (n); (a)++) {
    scanf("%d", &x[a]);
    if (ls[x[a]]) {
      nx[ls[x[a]]] = a;
      pr[a] = ls[x[a]];
    }
    ls[x[a]] = a;
  }
  int m;
  scanf("%d", &m);
  for (int(a) = (1); (a) <= (m); (a)++) {
    int l, r;
    scanf("%d%d", &l, &r);
    q[a] = query(l, r, a);
  }
  sort(q + 1, q + m + 1, cf);
  int fl = 1;
  int fr = 0;
  for (int(a) = (1); (a) <= (m); (a)++) {
    int cl = q[a].l;
    int cr = q[a].r;
    while (cl < fl) {
      fl--;
      add(fl);
      if (nx[fl] && nx[fl] <= fr) {
        if (dif[x[fl]][nx[fl] - fl] == 0) {
          if (nyala[x[fl]] == 1) can--;
          nyala[x[fl]]++;
          if (nyala[x[fl]] == 1) can++;
        }
        dif[x[fl]][nx[fl] - fl]++;
      }
    }
    while (cl > fl) {
      rem(fl);
      if (nx[fl] && nx[fl] <= fr) {
        if (dif[x[fl]][nx[fl] - fl] == 1) {
          if (nyala[x[fl]] == 1) can--;
          nyala[x[fl]]--;
          if (nyala[x[fl]] == 1) can++;
        }
        dif[x[fl]][nx[fl] - fl]--;
      }
      fl++;
    }
    while (cr > fr) {
      fr++;
      add(fr);
      if (pr[fr] && pr[fr] >= fl) {
        if (dif[x[fr]][fr - pr[fr]] == 0) {
          if (nyala[x[fr]] == 1) can--;
          nyala[x[fr]]++;
          if (nyala[x[fr]] == 1) can++;
        }
        dif[x[fr]][fr - pr[fr]]++;
      }
    }
    while (cr < fr) {
      rem(fr);
      if (pr[fr] && pr[fr] >= fl) {
        if (dif[x[fr]][fr - pr[fr]] == 1) {
          if (nyala[x[fr]] == 1) can--;
          nyala[x[fr]]--;
          if (nyala[x[fr]] == 1) can++;
        }
        dif[x[fr]][fr - pr[fr]]--;
      }
      fr--;
    }
    if (single > 0 || can > 0)
      ans[q[a].id] = on;
    else
      ans[q[a].id] = on + 1;
  }
  for (int(a) = (1); (a) <= (m); (a)++) {
    printf("%d\n", ans[a]);
  }
}
