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 MAXN = 100001;
const int SQRT = 317;
struct query {
  int l, r, id;
} que[MAXN];
bool compareMOZ(query a, query b) {
  if (a.l / SQRT != b.l / SQRT) return a.l / SQRT < b.l / SQRT;
  return a.r < b.r;
}
int a[MAXN];
int n, lef, rig;
int cnt[MAXN], dif, good;
int ans[MAXN], bad[MAXN];
deque<int> arr[MAXN];
bool is_bad(int x, int y) {
  return arr[x][y + 1] - arr[x][y] != arr[x][y] - arr[x][y - 1];
}
void add(int idx) {
  int val = a[idx];
  if (cnt[val] == 0) dif++;
  cnt[val]++;
  int sz = arr[val].size() + 1;
  if (sz == 1)
    arr[val].push_back(idx), good++;
  else if (arr[val][0] > idx) {
    arr[val].push_front(idx);
    if (sz > 2 && is_bad(val, 1)) {
      if (bad[val] == 0) good--;
      bad[val]++;
    }
  } else {
    arr[val].push_back(idx);
    if (sz > 2 && is_bad(val, sz - 2)) {
      if (bad[val] == 0) good--;
      bad[val]++;
    }
  }
}
void era(int idx) {
  int val = a[idx];
  cnt[val]--;
  if (cnt[val] == 0) dif--;
  int sz = arr[val].size();
  if (*arr[val].begin() == idx) {
    if (sz > 2 && is_bad(val, 1)) {
      if (bad[val] == 1) good++;
      bad[val]--;
    } else if (sz == 1)
      good--;
    arr[val].pop_front();
  } else {
    if (sz > 2 && is_bad(val, sz - 2)) {
      if (bad[val] == 1) good++;
      bad[val]--;
    }
    if (sz == 1) good--;
    arr[val].pop_back();
  }
}
void move(int x) {
  while (lef > que[x].l) add(--lef);
  while (rig < que[x].r) add(rig++);
  while (lef < que[x].l) era(lef++);
  while (rig > que[x].r) era(--rig);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> que[i].l >> que[i].r;
    que[i].l--;
    que[i].id = i;
  }
  sort(que, que + q, compareMOZ);
  for (int i = 0; i < q; ++i) {
    move(i);
    ans[que[i].id] = dif - (good > 0) + 1;
  }
  for (int i = 0; i < q; ++i) cout << ans[i] << "\n";
  return 0;
}
