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;
int n, q, arr[200005], ans[200005];
int curL = 0, curR = 0;
int cnt[200005], dist = 0;
deque<int> V[200005];
map<int, int> mp[200005];
int tot[200005];
int init = 0;
struct node {
  int l, r, idx;
} qus[200005];
bool comp(node n1, node n2) {
  if (n1.l / 310 != n2.l / 310) return n1.l / 310 < n2.l / 310;
  return n1.r < n2.r;
}
void add(int pos, bool f) {
  if (pos == 0) return;
  cnt[arr[pos]]++;
  if (cnt[arr[pos]] == 1) dist++;
  int val = arr[pos];
  int sz = V[val].size();
  if (f) {
    if (sz) {
      int diff = pos - V[val].back();
      mp[val][diff]++;
      if (mp[val][diff] == 1) {
        tot[val]++;
        if (tot[val] == 2) {
          init++;
        }
      }
    }
    V[val].push_back(pos);
  } else {
    if (sz) {
      int diff = V[val].front() - pos;
      mp[val][diff]++;
      if (mp[val][diff] == 1) {
        tot[val]++;
        if (tot[val] == 2) {
          init++;
        }
      }
    }
    V[val].push_front(pos);
  }
}
void dlt(int pos, bool f) {
  if (pos == 0) return;
  cnt[arr[pos]]--;
  if (cnt[arr[pos]] == 0) dist--;
  int val = arr[pos];
  if (f) {
    int last = V[val].back();
    V[val].pop_back();
    int sz = V[val].size();
    if (sz) {
      int diff = last - V[val].back();
      mp[val][diff]--;
      if (mp[val][diff] == 0) {
        tot[val]--;
        if (tot[val] == 1) init--;
      }
    }
  } else {
    int first = V[val].front();
    V[val].pop_front();
    int sz = V[val].size();
    if (sz) {
      int diff = V[val].front() - first;
      mp[val][diff]--;
      if (mp[val][diff] == 0) {
        tot[val]--;
        if (tot[val] == 1) init--;
      }
    }
  }
}
int main() {
  memset(cnt, 0, sizeof cnt);
  memset(tot, 0, sizeof tot);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    qus[i] = {l, r, i};
  }
  sort(qus + 1, qus + q + 1, comp);
  for (int i = 1; i <= q; i++) {
    int l = qus[i].l, r = qus[i].r;
    while (curR < r) {
      curR++;
      add(curR, true);
    }
    while (curR > r) {
      dlt(curR, true);
      curR--;
    }
    while (curL < l) {
      dlt(curL, false);
      curL++;
    }
    while (curL > l) {
      curL--;
      add(curL, false);
    }
    int id = qus[i].idx;
    int sum = dist;
    if (dist == init) sum++;
    ans[id] = sum;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
