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;
int II() {
  int q;
  scanf("%d", &q);
  return q;
}
long long LII() {
  long long q;
  scanf("%lld", &q);
  return q;
}
const int Mx = 1000005;
const int INF = 1e9 + 7;
int N;
int ar[Mx], L[Mx], is[Mx], pre[Mx];
struct info {
  int mx, cnt;
};
info tree[Mx * 4];
info combine(info l, info r) { return {min(l.mx, r.mx), l.cnt + r.cnt}; }
void build(int node, int b, int e) {
  tree[node] = {INF, 0};
  if (b == e) {
    tree[node] = {INF, 0};
  } else {
    int mid = (b + e) / 2;
    build(node * 2, b, mid);
    build(node * 2 + 1, mid + 1, e);
  }
}
info query(int node, int b, int e, int l, int r) {
  if (l > e or r < b) return {INF, 0};
  if (b >= l and e <= r) return tree[node];
  int mid = (b + e) / 2;
  return combine(query(node * 2, b, mid, l, r),
                 query(node * 2 + 1, mid + 1, e, l, r));
}
void update(int node, int b, int e, int in, info P) {
  if (in < b or in > e) return;
  if (b == e && in == e) {
    tree[node] = P;
  } else {
    int mid = (b + e) / 2;
    update(node * 2, b, mid, in, P);
    update(node * 2 + 1, mid + 1, e, in, P);
    tree[node] = combine(tree[node * 2], tree[node * 2 + 1]);
  }
}
void PreCalc() {
  fill(L, L + Mx, -INF);
  for (int i = 1; i <= N; i++) {
    pre[i] = L[ar[i]], L[ar[i]] = i;
  }
  for (int i = 1; i <= N; i++) {
    if (pre[i] == -INF || pre[pre[i]] == -INF)
      L[i] = -INF;
    else if (i - pre[i] == pre[i] - pre[pre[i]])
      L[i] = L[pre[i]];
    else
      L[i] = pre[pre[i]];
  }
}
vector<pair<int, int> > v[Mx];
void solve() {
  N = II();
  for (int i = 1; i <= N; i++) ar[i] = II();
  PreCalc();
  build(1, 1, N);
  int q;
  q = II();
  for (int i = 1; i <= q; i++) {
    int l, r;
    l = II(), r = II();
    v[r].push_back({l, i});
  }
  vector<pair<int, int> > ans;
  for (int i = 1; i <= N; i++) {
    if (pre[i] == -INF)
      update(1, 1, N, i, {L[i], 1});
    else
      update(1, 1, N, pre[i], {INF, 0}), update(1, 1, N, i, {L[i], 1});
    for (auto it : v[i]) {
      info Q = query(1, 1, N, it.first, i);
      if (Q.mx >= it.first) Q.cnt++;
      ans.push_back({it.second, Q.cnt});
    }
  }
  sort(ans.begin(), ans.end());
  for (auto it : ans) {
    printf("%d\n", it.second);
  }
}
int32_t main() {
  int tt;
  tt = 1;
  int cas = 1;
  while (tt--) {
    solve();
  }
}
