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 N = 100005;
vector<int> qs[N];
int b[N], q[N][2];
int s[8 * N], e[8 * N], d[8 * N];
int lastpos, etmp = 0, now, dtmp;
map<int, int> last;
void add(int x, int L, int R, int qx, int qv) {
  if (L == R) {
    s[x] += qv;
    if (qv == -1) {
      etmp = e[x];
      dtmp = d[x];
      e[x] = N;
    } else {
      if (lastpos < 0)
        d[x] = 0;
      else
        d[x] = now - lastpos;
      if (dtmp == now - lastpos || dtmp == 0)
        e[x] = etmp;
      else
        e[x] = lastpos - dtmp + 1;
    }
  } else {
    int M = (L + R) / 2;
    if (qx <= M)
      add(x * 2, L, M, qx, qv);
    else
      add(x * 2 + 1, M + 1, R, qx, qv);
    s[x] = s[x * 2] + s[x * 2 + 1];
    e[x] = min(e[x * 2], e[x * 2 + 1]);
  }
}
int ask(int x, int L, int R, int qx) {
  if (L == R) return s[x];
  int M = (L + R) / 2;
  if (qx <= M)
    return ask(x * 2, L, M, qx) + s[x * 2 + 1];
  else
    return ask(x * 2 + 1, M + 1, R, qx);
}
int askmin(int x, int L, int R, int qx) {
  if (L == R) return e[x];
  int M = (L + R) / 2;
  if (qx <= M)
    return min(askmin(x * 2, L, M, qx), e[x * 2 + 1]);
  else
    return askmin(x * 2 + 1, M + 1, R, qx);
}
void trace(int x, int L, int R) {
  printf("%d: [%d,%d] (%d,%d,%d)\n", x, L, R, s[x], e[x], d[x]);
  if (L == R) return;
  trace(x * 2, L, (L + R) / 2);
  trace(x * 2 + 1, (L + R) / 2 + 1, R);
}
int qans[N];
int main(void) {
  int m, qq;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
  scanf("%d", &qq);
  for (int i = 0; i < qq; i++) {
    scanf("%d%d", &q[i][0], &q[i][1]);
    qs[q[i][1]].push_back(i);
  }
  for (int i = 0; i < 4 * N; i++) e[i] = N;
  for (int i = 1; i <= m; i++) {
    if (last.count(b[i])) {
      int p = last[b[i]];
      etmp = 0;
      dtmp = 0;
      lastpos = p;
      now = i;
      add(1, 1, m, p, -1);
      add(1, 1, m, i, 1);
      last[b[i]] = i;
    } else {
      etmp = 0;
      dtmp = 0;
      lastpos = -10000;
      now = i;
      add(1, 1, m, i, 1);
      last[b[i]] = i;
    }
    for (__typeof((qs[i]).begin()) it = (qs[i]).begin(); it != (qs[i]).end();
         it++) {
      int l = q[*it][0], ans = 0;
      ans = ask(1, 1, m, l) + (askmin(1, 1, m, l) > l);
      qans[*it] = ans;
    }
  }
  for (int i = 0; i < qq; i++) printf("%d\n", qans[i]);
  return 0;
}
