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 INF = 1e9 + 250;
long long mod = 1e9 + 7;
const int N = 100005;
int a[N];
struct Node {
  int l, r, id;
};
Node qu[N];
bool cmp(const Node &p1, const Node &p2) { return p1.r < p2.r; }
int c[N];
int lowbit(int x) { return x & -x; }
int sum(int x) {
  int ret = 0;
  while (x > 0) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
void add(int x, int d, int n) {
  while (x <= n) {
    c[x] += d;
    x += lowbit(x);
  }
}
int e[N];
int sum2(int x) {
  int ret = 0;
  while (x > 0) {
    ret += e[x];
    x -= lowbit(x);
  }
  return ret;
}
void add2(int x, int d, int n) {
  while (x <= n) {
    e[x] += d;
    x += lowbit(x);
  }
}
int ans[N];
int last[N];
int b[N];
int d[N];
int f[N];
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    b[i] = last[a[i]];
    last[a[i]] = i;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    qu[i].id = i;
    scanf("%d%d", &qu[i].l, &qu[i].r);
  }
  sort(qu + 1, qu + q + 1, cmp);
  int pos = 1;
  for (int i = 1; i <= q; i++) {
    while (pos <= qu[i].r) {
      add2(pos, 1, n);
      if (b[pos] != 0) add2(b[pos], -1, n);
      int t;
      if (b[pos] == 0) {
        t = 0;
        add(pos, 1, n);
      } else {
        t = pos - b[pos];
        add(pos, 1, n);
        add(b[pos], -1, n);
        if (t != d[a[pos]]) {
          if (f[a[pos]] != 0) add(f[a[pos]], 1, n);
          int x = b[b[pos]];
          if (x != 0) {
            add(x, -1, n);
            f[a[pos]] = x;
          }
          d[a[pos]] = t;
        }
      }
      pos++;
    }
    ans[qu[i].id] = sum2(qu[i].r) - sum2(qu[i].l - 1) +
                    (sum(qu[i].r) - sum(qu[i].l - 1) == 0);
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
