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;
struct str {
  int l, r, i;
} a[100002];
int t[100002];
int yo[100002];
int p[100002], f[100002];
int ans[100002];
bool bj[100002];
int n, m, num;
int sum;
int sum1;
int l = 1, r = 0;
int v[100002], b[100002];
bool operator<(const str &l, const str &r) {
  if (b[l.l] ^ b[r.l]) return l.l < r.l;
  if (b[l.l] & 1) return l.r > r.r;
  return l.r < r.r;
}
void addr(int x) {
  if (!t[v[x]]) {
    sum++;
    sum1++;
    bj[v[x]] = 1;
  }
  if (f[x] > l) {
    if (bj[v[x]]) {
      bj[v[x]] = 0;
      sum1--;
    }
  }
  yo[v[x]] = x;
  t[v[x]]++;
}
void delr(int x) {
  yo[v[x]] = p[x];
  t[v[x]]--;
  if (!t[v[x]]) {
    sum--;
    if (bj[v[x]]) sum1--;
    bj[v[x]] = 0;
  } else {
    if (f[p[x]] <= l && !bj[v[x]]) {
      bj[v[x]] = 1;
      sum1++;
    }
  }
}
void addl(int x) {
  if (!t[v[x]]) {
    sum++;
    sum1++;
    bj[v[x]] = 1;
    yo[v[x]] = x;
  }
  if (f[yo[v[x]]] > x && bj[v[x]]) {
    bj[v[x]] = 0;
    sum1--;
  }
  if (f[yo[v[x + 1]]] > x && bj[v[x + 1]]) {
    bj[v[x + 1]] = 0;
    sum1--;
  }
  t[v[x]]++;
}
void dell(int x) {
  t[v[x]]--;
  if (!t[v[x]]) {
    sum--;
    if (bj[v[x]]) sum1--;
    bj[v[x]] = 0;
  } else if (f[yo[v[x]]] <= x + 1 && !bj[v[x]]) {
    bj[v[x]] = 1;
    sum1++;
  }
  if (f[yo[v[x + 1]]] <= x + 1 && !bj[v[x + 1]]) {
    bj[v[x + 1]] = 1;
    sum1++;
  }
}
int main() {
  cin >> n;
  num = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    b[i] = (i - 1) / num;
  }
  for (int i = 1; i <= n; i++) {
    if (!yo[v[i]]) {
      p[i] = 0;
      f[i] = 0;
      yo[v[i]] = i;
    } else {
      p[i] = yo[v[i]];
      if (p[yo[v[i]]] == 0) {
        f[i] = 0;
      } else {
        if (yo[v[i]] - p[yo[v[i]]] == i - yo[v[i]]) {
          f[i] = f[yo[v[i]]];
        } else {
          f[i] = p[p[i]] + 1;
        }
      }
      yo[v[i]] = i;
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].i = i;
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) {
    while (r < a[i].r) addr(++r);
    while (l > a[i].l) addl(--l);
    while (r > a[i].r) delr(r--);
    while (l < a[i].l) dell(l++);
    ans[a[i].i] = sum;
    if (!sum1) {
      ans[a[i].i]++;
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}
