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 M = (int)1e5 + 5;
int n;
int a[M];
int q, ans[M];
vector<pair<int, int> > ps[M];
int pas[M], las[M], llas[M], mfr[M];
struct SEGMENT1 {
  int val[M * 4];
  void update(int L, int R, int p, int l, int r, int v) {
    if (L == l && R == r) {
      val[p] += v;
      return;
    }
    int mid = (L + R) >> 1;
    if (mid >= r)
      update(L, mid, p << 1, l, r, v);
    else if (mid < l)
      update(mid + 1, R, p << 1 | 1, l, r, v);
    else
      update(L, mid, p << 1, l, mid, v),
          update(mid + 1, R, p << 1 | 1, mid + 1, r, v);
  }
  int query(int L, int R, int p, int x) {
    if (L == R) return val[p];
    int mid = (L + R) >> 1;
    if (mid >= x)
      return val[p] + query(L, mid, p << 1, x);
    else
      return val[p] + query(mid + 1, R, p << 1 | 1, x);
  }
} T1;
struct SEGMENT2 {
  int val[M * 4];
  void update(int L, int R, int p, int x, int v) {
    val[p] += v;
    if (L == R) return;
    int mid = (L + R) >> 1;
    if (mid >= x)
      update(L, mid, p << 1, x, v);
    else
      update(mid + 1, R, p << 1 | 1, x, v);
  }
  int query(int L, int R, int p, int l, int r) {
    if (L == l && R == r) return val[p];
    int mid = (L + R) >> 1;
    if (mid >= r)
      return query(L, mid, p << 1, l, r);
    else if (mid < l)
      return query(mid + 1, R, p << 1 | 1, l, r);
    else
      return query(L, mid, p << 1, l, mid) +
             query(mid + 1, R, p << 1 | 1, mid + 1, r);
  }
} T2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &q);
  for (int i = 1, l, r; i <= q; i++)
    scanf("%d %d", &l, &r), ps[r].push_back(pair<int, int>(l, i));
  for (int i = 1, x, len; i <= n; i++) {
    x = a[i];
    len = i - las[x];
    if (len != pas[x] && pas[x] != las[x]) {
      if (mfr[x] < llas[x]) T1.update(1, n, 1, mfr[x] + 1, llas[x], 1);
      mfr[x] = llas[x];
    }
    pas[x] = len;
    if (las[x] != 0) T2.update(1, n, 1, las[x], -1);
    T2.update(1, n, 1, i, 1);
    llas[x] = las[x];
    las[x] = i;
    for (int j = 0, up = ps[i].size(), c1, c2, l, id; j < up; j++) {
      l = ps[i][j].first;
      id = ps[i][j].second;
      c1 = T2.query(1, n, 1, l, i);
      c2 = T1.query(1, n, 1, l);
      if (c1 != c2)
        ans[id] = c1;
      else
        ans[id] = c1 + 1;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
