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 sz = 325;
struct node {
  int b, r, i;
};
bool cmp(node a, node b) {
  if (a.b != b.b) return a.b < b.b;
  return a.r < b.r;
}
node block[100005];
int a[100005], l[100005], r[100005], ans[100005];
int last[100005], pv[100005], nt[100005];
deque<int> dq[100005];
int diff[100005], x = 0;
int cnt[100005], d = 0;
int curl = 1, curr = 0;
void add(int idx) {
  if (curl <= pv[idx] && pv[idx] <= curr) {
    if (!dq[a[idx]].empty() && dq[a[idx]].back() != idx - pv[idx]) {
      diff[a[idx]]++;
      if (diff[a[idx]] == 1) x++;
    }
    dq[a[idx]].push_back(idx - pv[idx]);
  }
  if (curl <= nt[idx] && nt[idx] <= curr) {
    if (!dq[a[idx]].empty() && dq[a[idx]].front() != nt[idx] - idx) {
      diff[a[idx]]++;
      if (diff[a[idx]] == 1) x++;
    }
    dq[a[idx]].push_front(nt[idx] - idx);
  }
  cnt[a[idx]]++;
  if (cnt[a[idx]] == 1) d++;
}
void del(int idx) {
  if (curl <= pv[idx] && pv[idx] <= curr) {
    if ((int)(dq[a[idx]]).size() > 1 &&
        dq[a[idx]][(int)(dq[a[idx]]).size() - 1] !=
            dq[a[idx]][(int)(dq[a[idx]]).size() - 2]) {
      diff[a[idx]]--;
      if (diff[a[idx]] == 0) x--;
    }
    dq[a[idx]].pop_back();
  }
  if (curl <= nt[idx] && nt[idx] <= curr) {
    if ((int)(dq[a[idx]]).size() > 1 && dq[a[idx]][0] != dq[a[idx]][1]) {
      diff[a[idx]]--;
      if (diff[a[idx]] == 0) x--;
    }
    dq[a[idx]].pop_front();
  }
  cnt[a[idx]]--;
  if (cnt[a[idx]] == 0) d--;
}
int main() {
  memset(pv, -1, sizeof(pv));
  memset(nt, -1, sizeof(nt));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (last[a[i]] != 0) pv[i] = last[a[i]];
    last[a[i]] = i;
  }
  memset(last, 0, sizeof(last));
  for (int i = n; i >= 1; i--) {
    if (last[a[i]] != 0) nt[i] = last[a[i]];
    last[a[i]] = i;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &l[i], &r[i]);
    block[i] = {l[i] / sz, r[i], i};
  }
  sort(block + 1, block + 1 + q, cmp);
  for (int i = 1; i <= q; i++) {
    int idx = block[i].i;
    while (curl < l[idx]) del(curl++);
    while (curl > l[idx]) add(--curl);
    while (curr < r[idx]) add(++curr);
    while (curr > r[idx]) del(curr--);
    ans[idx] = d;
    if (d == x) ans[idx]++;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
