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 MAXN = 1e5 + 100;
vector<int> V[MAXN];
int a[MAXN], FL[MAXN], FR[MAXN], lft[MAXN], rgt[MAXN], ans[MAXN];
int fen[MAXN];
int n;
bool c[MAXN];
void add(int p, int x) {
  p++;
  if (!p) {
    return;
  }
  for (p; p <= n; p += (p & -p)) {
    fen[p] += x;
  }
  return;
}
int ask(int p) {
  p++;
  int res = 0;
  for (p; p > 0; p -= (p & -p)) {
    res += fen[p];
  }
  return res;
}
void cler() {
  for (int i = 0; i <= n; i++) {
    fen[i] = 0;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    V[a[i]].push_back(i);
  }
  for (int i = 0; i < MAXN; i++) {
    int p = V[i].size() - 1;
    for (int j = V[i].size() - 1; j >= 0; j--) {
      int t = V[i][j + 1] - V[i][j];
      while (V[i][j] + (t * (p - j)) != V[i][p]) {
        p--;
      }
      FL[V[i][j]] = V[i][j - 1];
      FR[V[i][j]] = V[i][p + 1];
      if (j == 0) {
        FL[V[i][j]] = -1;
      }
      if (p == V[i].size() - 1) {
        FR[V[i][j]] = n;
      }
    }
  }
  vector<pair<int, int> > Q, W;
  for (int i = 0; i < n; i++) {
    W.push_back({FR[i], i});
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &lft[i], &rgt[i]);
    lft[i]--;
    rgt[i]--;
    Q.push_back({rgt[i], i});
  }
  sort(Q.begin(), Q.end());
  sort(W.begin(), W.end());
  int p = 0;
  int h = 0;
  for (int i = 0; i < n; i++) {
    add(FL[i], -1);
    add(i, 1);
    while (W[p].first == i) {
      int e = W[p].second;
      add(FL[e], 1);
      add(e, -1);
      p++;
    }
    while (Q[h].first == i) {
      int e = Q[h].second;
      c[e] = 1;
      if (ask(i) - ask(lft[e] - 1) >= 1) {
        c[e] = 0;
      }
      h++;
    }
  }
  cler();
  h = 0;
  for (int i = 0; i < n; i++) {
    add(FL[i], -1);
    add(i, 1);
    while (Q[h].first == i) {
      int e = Q[h].second;
      ans[e] = ask(i) - ask(lft[e] - 1);
      h++;
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i] + c[i]);
  }
  return 0;
}
