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 {
  int l, r, num;
} ask[100005];
deque<int> q[100005];
int n, m, sq, sum, sumk, a[100005], bl[100005], prt[100005], s[100005],
    ans[100005];
bool vst[100005];
bool operator<(const node &a, const node &b) {
  if (bl[a.l] ^ bl[b.l]) return a.l < b.l;
  if (bl[a.l] & 1) return a.r < b.r;
  return a.r > b.r;
}
int Getf(int x) {
  if (x == prt[x]) return prt[x];
  return prt[x] = Getf(prt[x]);
}
void Addl(int x) {
  s[a[x]]++;
  if (s[a[x]] == 1) {
    sum++;
    sumk++;
    vst[a[x]] = true;
  } else if (s[a[x]] != 2) {
    if (q[a[x]].front() - x != q[a[x]][1] - q[a[x]][0]) {
      if (vst[a[x]]) {
        vst[a[x]] = false;
        sumk--;
      }
    } else
      prt[x] = Getf(q[a[x]].front());
  }
  q[a[x]].push_front(x);
}
void Addr(int x) {
  s[a[x]]++;
  if (s[a[x]] == 1) {
    sum++;
    sumk++;
    vst[a[x]] = true;
  } else if (s[a[x]] != 2) {
    if (x - q[a[x]].back() != q[a[x]][s[a[x]] - 2] - q[a[x]][s[a[x]] - 3]) {
      if (vst[a[x]]) {
        vst[a[x]] = false;
        sumk--;
      }
    } else
      prt[q[a[x]][s[a[x]] - 3]] = q[a[x]][s[a[x]] - 2];
  }
  q[a[x]].push_back(x);
}
void Dell(int x) {
  s[a[x]]--;
  q[a[x]].pop_front();
  prt[x] = x;
  if (!s[a[x]]) {
    sum--;
    sumk--;
    vst[a[x]] = false;
  } else {
    if (!vst[a[x]]) {
      if ((s[a[x]] == 1 || s[a[x]] == 2)) {
        vst[a[x]] = true;
        sumk++;
      } else if (Getf(q[a[x]].front()) == Getf(q[a[x]][s[a[x]] - 2])) {
        vst[a[x]] = true;
        sumk++;
      }
    }
  }
}
void Delr(int x) {
  s[a[x]]--;
  q[a[x]].pop_back();
  if (!s[a[x]]) {
    sum--;
    sumk--;
    vst[a[x]] = false;
  } else {
    if (!vst[a[x]]) {
      if (s[a[x]] == 1 || s[a[x]] == 2) {
        vst[a[x]] = true;
        sumk++;
      } else {
        if (Getf(q[a[x]].front()) == Getf(q[a[x]][s[a[x]] - 2])) {
          vst[a[x]] = true;
          sumk++;
        }
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  sq = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    bl[i] = (i - 1) / sq + 1;
    prt[i] = i;
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &ask[i].l, &ask[i].r);
    ask[i].num = i;
  }
  sort(ask + 1, ask + m + 1);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (l > ask[i].l) Addl(--l);
    while (r < ask[i].r) Addr(++r);
    while (l < ask[i].l) Dell(l++);
    while (r > ask[i].r) Delr(r--);
    ans[ask[i].num] = sum;
    if (!sumk) ans[ask[i].num]++;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
