Ildar is the algorithm teacher of William and Harris. Today, Ildar is teaching Cartesian Tree. However, Harris is sick, so Ildar is only teaching William.

A cartesian tree is a rooted tree, that can be constructed from a sequence of distinct integers. We build the cartesian tree as follows:

  1. If the sequence is empty, return an empty tree;
  2. Let the position of the maximum element be x;
  3. Remove element on the position x from the sequence and break it into the left part and the right part (which might be empty) (not actually removing it, just taking it away temporarily);
  4. Build cartesian tree for each part;
  5. Create a new vertex for the element, that was on the position x which will serve as the root of the new tree. Then, for the root of the left part and right part, if exists, will become the children for this vertex;
  6. Return the tree we have gotten.



For example, this is the cartesian tree for the sequence 4, 2, 7, 3, 5, 6, 1:

<image>

After teaching what the cartesian tree is, Ildar has assigned homework. He starts with an empty sequence a.

In the i-th round, he inserts an element with value i somewhere in a. Then, he asks a question: what is the sum of the sizes of the subtrees for every node in the cartesian tree for the current sequence a?

Node v is in the node u subtree if and only if v = u or v is in the subtree of one of the vertex u children. The size of the subtree of node u is the number of nodes v such that v is in the subtree of u.

Ildar will do n rounds in total. The homework is the sequence of answers to the n questions.

The next day, Ildar told Harris that he has to complete the homework as well. Harris obtained the final state of the sequence a from William. However, he has no idea how to find the answers to the n questions. Help Harris!

Input

The first line contains a single integer n (1 ≤ n ≤ 150000).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guarenteed that each integer from 1 to n appears in the sequence exactly once.

Output

Print n lines, i-th line should contain a single integer — the answer to the i-th question.

Examples

Input


5
2 4 1 5 3


Output


1
3
6
8
11


Input


6
1 2 4 5 6 3


Output


1
3
6
8
12
17

Note

After the first round, the sequence is 1. The tree is

<image>

The answer is 1.

After the second round, the sequence is 2, 1. The tree is

<image>

The answer is 2+1=3.

After the third round, the sequence is 2, 1, 3. The tree is

<image>

The answer is 2+1+3=6.

After the fourth round, the sequence is 2, 4, 1, 3. The tree is

<image>

The answer is 1+4+1+2=8.

After the fifth round, the sequence is 2, 4, 1, 5, 3. The tree is

<image>

The answer is 1+3+1+5+1=11.
#include <bits/stdc++.h>
using namespace std;
int n, sq, s;
vector<long long int> ans(200005, 0);
long long int answ = 0;
struct Block {
  vector<pair<int, list<int>>> R;
  int size = 0;
  long long int sum = 0;
  int lazy = 0;
  void up(long long int sign) { answ += sign * sum; }
  void add(int i, int r, int pt) {
    up(-1);
    sum = 0;
    vector<pair<int, int>> left, right;
    for (auto& p : R) {
      for (auto q : p.second) {
        if (q < i)
          left.push_back({min(p.first + lazy, pt), q});
        else
          right.push_back({p.first + lazy + 1, q});
      }
    }
    R.clear();
    left.push_back({r, i});
    size++;
    lazy = 0;
    int a = 0;
    for (auto p : left) {
      while (a != ((int)(right).size()) && p.first > right[a].first) {
        if (R.empty() || R.back().first != right[a].first) {
          R.push_back({right[a].first, list<int>()});
        }
        R.back().second.push_back(right[a].second);
        a++;
      }
      if (R.empty() || R.back().first != p.first) {
        R.push_back({p.first, list<int>()});
      }
      R.back().second.push_back(p.second);
    }
    while (a != ((int)(right).size())) {
      if (R.empty() || R.back().first != right[a].first) {
        R.push_back({right[a].first, list<int>()});
      }
      R.back().second.push_back(right[a].second);
      a++;
    }
    for (auto& p : R) {
      sum += p.first * ((int)(p.second).size());
    }
    up(1);
  }
  void update(int r) {
    r -= lazy;
    if (R.empty() || R.back().first <= r) return;
    up(-1);
    sum -= (R.back().first - r) * ((int)(R.back().second).size());
    R.back().first = r;
    for (int i = ((int)(R).size()) - 2; i >= 0 && R[i].first >= r; i--) {
      sum -= (R[i].first - r) * ((int)(R[i].second).size());
      R[i].second.splice(R[i].second.end(), R[i + 1].second);
      R[i].first = r;
      R.pop_back();
    }
    up(1);
  }
  void update() {
    up(-1);
    lazy++;
    sum += size;
    up(1);
  }
};
struct Fenwick {
  vector<int> bit;
  int size;
  void modify(int j, int v) {
    j++;
    for (; j < size; j += j & -j) bit[j] += v;
  }
  int query(int j) {
    j++;
    int v = 0;
    for (; j > 0; j -= j & -j) v += bit[j];
    return v;
  }
  Fenwick(int s) {
    s += 3;
    size = s;
    bit = vector<int>(s, 0);
  }
};
void solve(vector<int>& arr, int sign) {
  vector<int> ptr(n + 1);
  for (int i = 0; i < n; i++) ptr[arr[i]] = i;
  Block data[s];
  Fenwick cnt(n);
  answ = 0;
  for (int i = 1; i <= n; i++) {
    int pt = cnt.query(ptr[i]);
    cnt.modify(ptr[i], 1);
    for (int b = 0; b < s; b++) {
      if (b < ptr[i] / sq) {
        data[b].update(pt);
      }
      if (b == ptr[i] / sq) {
        data[b].add(ptr[i], i, pt);
      }
      if (b > ptr[i] / sq) {
        data[b].update();
      }
    }
    if (sign > 0)
      ans[i] += answ;
    else
      ans[i] -= 1ll * i * i - answ;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  sq = 200;
  s = (n - 1) / sq + 1;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  solve(arr, 1);
  reverse((arr).begin(), (arr).end());
  solve(arr, -1);
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
}
