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 N = 1e5 + 5;
int unit, ans, dc;
int a[N], fl[N], fr[N], pre[N], bak[N], last[N], res[N], cnt[N];
int id;
struct node {
  int l, r;
  int id;
} q[N];
bool cmp(node a, node b) {
  return a.l / unit == b.l / unit ? a.r < b.r : a.l / unit < b.l / unit;
}
void addl(int pos, int r) {
  cnt[a[pos]]++;
  if (cnt[a[pos]] == 1) {
    ans++;
    dc++;
  } else if (fr[pos] <= r && fr[bak[pos]] > r)
    dc--;
}
void removel(int pos, int r) {
  cnt[a[pos]]--;
  if (cnt[a[pos]] == 0) {
    ans--;
    dc--;
  } else if (fr[bak[pos]] > r && fr[pos] <= r)
    dc++;
}
void addr(int pos, int l) {
  cnt[a[pos]]++;
  if (cnt[a[pos]] == 1) {
    ans++;
    dc++;
  } else if (fl[pos] >= l && fl[pre[pos]] < l)
    dc--;
}
void remover(int pos, int l) {
  cnt[a[pos]]--;
  if (cnt[a[pos]] == 0) {
    ans--;
    dc--;
  } else if (fl[pre[pos]] < l && fl[pos] >= l)
    dc++;
}
int main() {
  int n;
  scanf("%d", &n);
  unit = sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    pre[i] = last[a[i]];
    last[a[i]] = i;
    if (pre[pre[i]] == 0)
      fl[i] == 0;
    else if (pre[i] - pre[pre[i]] == i - pre[i])
      fl[i] = fl[pre[i]];
    else
      fl[i] = pre[pre[i]];
  }
  memset(last, 0, sizeof(last));
  for (int i = n; i >= 1; i--) {
    bak[i] = last[a[i]];
    last[a[i]] = i;
    if (bak[bak[i]] == 0)
      fr[i] = n + 1;
    else if (bak[i] - bak[bak[i]] == i - bak[i])
      fr[i] = fr[bak[i]];
    else
      fr[i] = bak[bak[i]];
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m, cmp);
  int L = q[1].l, R = L - 1;
  for (int i = 1; i <= m; i++) {
    while (L > q[i].l) {
      addl(--L, R);
    }
    while (L < q[i].l) {
      removel(L++, R);
    }
    while (R < q[i].r) {
      addr(++R, L);
    }
    while (R > q[i].r) {
      remover(R--, L);
    }
    if (dc > 0)
      res[q[i].id] = ans;
    else
      res[q[i].id] = ans + 1;
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", res[i]);
  }
  return 0;
}
