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>
#pragma GCC optimize("O3")
using namespace std;
const int sq = 320;
int brr[100005];
vector<pair<pair<int, int>, int> > qr[sq];
vector<int> pos[100005];
int nxt[100005], prv[100005];
int ans = 0;
int freq[100005];
int out[100005];
int sum = 0;
int dis[100005];
map<int, int> mp[100005];
void add(int x, int rt) {
  int y = brr[x];
  freq[y]++;
  if (freq[y] == 1) {
    ans++;
    sum++;
    return;
  }
  if (rt == 0) {
    int d = nxt[x] - x;
    if (mp[y].find(d) == mp[y].end()) mp[y][d] = 0;
    int z = (++mp[y][d]);
    if (z == 1) {
      dis[y]++;
      if (dis[y] == 2) sum--;
    }
  } else {
    int d = x - prv[x];
    if (mp[y].find(d) == mp[y].end()) mp[y][d] = 0;
    int z = (++mp[y][d]);
    if (z == 1) {
      dis[y]++;
      if (dis[y] == 2) sum--;
    }
  }
}
void rmv(int x, int rt) {
  int y = brr[x];
  freq[y]--;
  if (freq[y] == 0) {
    ans--;
    sum--;
    return;
  }
  if (rt == 0) {
    int d = nxt[x] - x;
    if (mp[y].find(d) == mp[y].end()) mp[y][d] = 0;
    int z = (--mp[y][d]);
    if (z == 0) {
      dis[brr[x]]--;
      if (dis[brr[x]] == 1) sum++;
    }
  } else {
    int d = x - prv[x];
    if (mp[y].find(d) == mp[y].end()) mp[y][d] = 0;
    int z = (--mp[y][d]);
    if (z == 0) {
      dis[brr[x]]--;
      if (dis[brr[x]] == 1) sum++;
    }
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> brr[i];
    pos[brr[i]].push_back(i);
  }
  for (int i = 1; i < 100005; i++) {
    for (int j = 0; j + 1 < pos[i].size(); j++) {
      nxt[pos[i][j]] = pos[i][j + 1];
    }
    if (pos[i].size() != 0) nxt[pos[i][pos[i].size() - 1]] = -1;
  }
  for (int i = 1; i < 100005; i++) {
    if (pos[i].size() != 0) prv[pos[i][0]] = -1;
    for (int j = 1; j < pos[i].size(); j++) {
      prv[pos[i][j]] = pos[i][j - 1];
    }
  }
  int q;
  cin >> q;
  int l, r;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    int z = l / sq;
    qr[z].push_back({{r, l}, i});
  }
  l = qr[0][0].first.second;
  r = l;
  freq[brr[l]]++;
  ans = 1;
  sum = 1;
  for (int i = 0; i < sq; i++) {
    sort(qr[i].begin(), qr[i].end());
    for (auto j : qr[i]) {
      int lf = j.first.second;
      int rt = j.first.first;
      while (lf < l) {
        l--;
        add(l, 0);
      }
      while (rt > r) {
        r++;
        add(r, 1);
      }
      while (lf > l) {
        rmv(l, 0);
        l++;
      }
      while (rt < r) {
        rmv(r, 1);
        r--;
      }
      out[j.second] = ans;
      if (sum == 0) out[j.second]++;
    }
  }
  for (int i = 1; i <= q; i++) {
    cout << out[i] << '\n';
  }
  return 0;
}
