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 NMAX = 100005;
int v[NMAX];
int sol[NMAX];
struct Query {
  int st, dr;
  int bucket;
  int poz;
};
Query q[NMAX];
bool cmp(Query first, Query second) {
  if (first.bucket == second.bucket) {
    return first.dr < second.dr;
  }
  return first.bucket < second.bucket;
}
int ant[NMAX];
int lst[NMAX];
int nxt[NMAX];
int dpst[NMAX];
int dpdr[NMAX];
int vf[NMAX];
int prg = 0, nr = 0;
int st = 0, dr = 0;
void addst(int p) {
  int x = v[p];
  if (vf[x] == 0) {
    nr++;
    prg++;
  } else if (dpdr[p] < dr && dr <= dpdr[nxt[p]]) {
    prg--;
  }
  vf[x]++;
}
void adddr(int p) {
  int x = v[p];
  if (vf[x] == 0) {
    nr++;
    prg++;
  } else if (dpst[p] > st && st >= dpst[lst[p]]) {
    prg--;
  }
  vf[x]++;
}
void erasest(int p) {
  int x = v[p];
  if (vf[x] == 1) {
    nr--;
    prg--;
  } else if (dpdr[p] < dr && dr <= dpdr[nxt[p]]) {
    prg++;
  }
  vf[x]--;
}
void erasedr(int p) {
  int x = v[p];
  if (vf[x] == 1) {
    nr--;
    prg--;
  } else if (dpst[p] > st && st >= dpst[lst[p]]) {
    prg++;
  }
  vf[x]--;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
  }
  for (int i = 1; i <= n; i++) {
    lst[i] = ant[v[i]];
    ant[v[i]] = i;
    if (lst[i] == 0) {
      dpst[i] = 0;
    } else if (lst[lst[i]] == 0) {
      dpst[i] = 0;
    } else if (i - lst[i] == lst[i] - lst[lst[i]]) {
      dpst[i] = dpst[lst[i]];
    } else {
      dpst[i] = lst[lst[i]] + 1;
    }
  }
  for (int i = 0; i <= NMAX - 2; i++) {
    ant[i] = 0;
  }
  for (int i = n; i > 0; i--) {
    nxt[i] = ant[v[i]];
    ant[v[i]] = i;
    if (nxt[i] == 0) {
      dpdr[i] = n + 1;
    } else if (nxt[nxt[i]] == 0) {
      dpdr[i] = n + 1;
    } else if (i - nxt[i] == nxt[i] - nxt[nxt[i]]) {
      dpdr[i] = dpdr[nxt[i]];
    } else {
      dpdr[i] = nxt[nxt[i]] - 1;
    }
  }
  scanf("%d", &m);
  int rad = (int)sqrt((double)n) + 1;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].st, &q[i].dr);
    q[i].poz = i;
    q[i].bucket = q[i].st / rad;
  }
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    while (dr < q[i].dr) {
      dr++;
      adddr(dr);
    }
    while (dr > q[i].dr) {
      erasedr(dr);
      dr--;
    }
    while (st < q[i].st) {
      erasest(st);
      st++;
    }
    while (st > q[i].st) {
      st--;
      addst(st);
    }
    sol[q[i].poz] = nr;
    if (prg == 0) {
      sol[q[i].poz]++;
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", sol[i]);
  }
  return 0;
}
