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 Node {
  Node *lc, *rc;
  int size, maxv;
} * Root[100010], *Null;
int n, a[100010], R[100010], last[100010];
int Log[100010], maxv[20][100010];
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first + b.first, max(a.second, b.second));
}
Node *Insert(Node *now, int l, int r, int x, int v) {
  Node *p = new Node;
  *p = *now;
  ++p->size, p->maxv = max(p->maxv, v);
  if (l == r) return p;
  int mid = (l + r) >> 1;
  if (x <= mid)
    p->lc = Insert(p->lc, l, mid, x, v);
  else
    p->rc = Insert(p->rc, mid + 1, r, x, v);
  return p;
}
pair<int, int> Query(Node *p, int l, int r, int x, int y) {
  if (l == x && r == y) return make_pair(p->size, p->maxv);
  int mid = (l + r) >> 1;
  if (y <= mid)
    return Query(p->lc, l, mid, x, y);
  else if (mid < x)
    return Query(p->rc, mid + 1, r, x, y);
  else
    return Query(p->lc, l, mid, x, mid) + Query(p->rc, mid + 1, r, mid + 1, y);
}
void read() {
  static pair<int, int> sorted[100010];
  static int now[100010], next[100010];
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = n; i >= 1; --i) next[i] = now[a[i]], now[a[i]] = i;
  for (int i = n; i >= 1; --i) {
    int x = next[i], y = next[x];
    if (y == 0)
      R[i] = n + 1;
    else
      R[i] = y - x == x - i ? R[x] : y;
    last[x] = i;
  }
  for (int i = 1; i <= n; ++i) sorted[i] = make_pair(last[i], i);
  sort(sorted + 1, sorted + n + 1);
  Null = new Node;
  Null->lc = Null, Null->rc = Null, Null->size = Null->maxv = 0;
  Root[0] = Null;
  int j = 1;
  while (sorted[j].first == 0 && j <= n)
    Root[0] = Insert(Root[0], 1, n, sorted[j].second, R[sorted[j].second]), ++j;
  for (int i = 1; i <= n + 1; ++i) {
    Root[i] = Root[i - 1];
    while (sorted[j].first == i)
      Root[i] = Insert(Root[i], 1, n, sorted[j].second, R[sorted[j].second]),
      ++j;
  }
}
void Query() {
  int m;
  scanf("%d", &m);
  for (int i = 1, l, r; i <= m; ++i) {
    scanf("%d %d", &l, &r);
    pair<int, int> ans = Query(Root[l - 1], 1, n, l, r);
    printf("%d\n", ans.first + (ans.second <= r));
  }
}
int main() {
  read();
  Query();
  return 0;
}
