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;
void setIO(const string &name) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin.exceptions(istream::failbit);
}
const int inf = 0x3f3f3f3f, mod = 1e9 + 7, maxn = 1e5 + 5, k = 300;
struct query {
  int l, r, ind;
  bool operator<(query other) const {
    if (l / k != other.l / k) {
      return l < other.l;
    }
    return (l / k & 1) ? (r < other.r) : (r > other.r);
  }
};
int cnt[maxn], cur = 0, a[maxn], good = 0;
deque<int> occ[maxn];
map<int, int> diff[maxn];
bool g[maxn];
void add(int x, bool r) {
  ++cnt[a[x]];
  if (cnt[a[x]] == 1) {
    ++good;
    g[a[x]] = true;
    ++cur;
  } else {
    int y;
    if (r) {
      y = occ[a[x]].back();
    } else {
      y = occ[a[x]].front();
    }
    ++diff[a[x]][abs(x - y)];
    if (diff[a[x]].size() >= 2) {
      if (g[a[x]]) {
        --good;
        g[a[x]] = false;
      }
    }
  }
  if (r) {
    occ[a[x]].push_back(x);
  } else {
    occ[a[x]].push_front(x);
  }
}
void remove(int x, bool r) {
  --cnt[a[x]];
  int y;
  if (r) {
    y = occ[a[x]].back();
    occ[a[x]].pop_back();
  } else {
    y = occ[a[x]].front();
    occ[a[x]].pop_front();
  }
  if (!cnt[a[x]]) {
    --good;
    --cur;
    g[a[x]] = false;
  } else {
    int z;
    if (r) {
      z = occ[a[x]].back();
    } else {
      z = occ[a[x]].front();
    }
    if (--diff[a[x]][abs(y - z)] == 0) {
      diff[a[x]].erase(abs(y - z));
      if (diff[a[x]].size() <= 1) {
        if (!g[a[x]]) {
          ++good;
          g[a[x]] = true;
        }
      }
    }
  }
}
int main() {
  setIO("2");
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int q;
  cin >> q;
  vector<query> queries(q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    queries[i] = {l, r, i};
  }
  sort(queries.begin(), queries.end());
  vector<long long> sol(q);
  int l = 0, r = -1;
  for (auto &i : queries) {
    while (r < i.r) {
      ++r;
      add(r, true);
    }
    while (l > i.l) {
      --l;
      add(l, false);
    }
    while (r > i.r) {
      remove(r, true);
      --r;
    }
    while (l < i.l) {
      remove(l, false);
      ++l;
    }
    sol[i.ind] = cur;
    if (!good) {
      ++sol[i.ind];
    }
  }
  for (auto &i : sol) {
    cout << i << "\n";
  }
  return 0;
}
