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;
const int N = 15e4 + 5;
int n, p[N], fen[N];
struct Segment {
  struct Node {
    long long sum, lz;
    int mx1, mx2, ted, k, lz_mx;
    Node() : sum(0), lz(0), mx1(-N), mx2(-N), ted(0), k(0), lz_mx(0){};
    void add(int x) {
      if (!k) return;
      sum += 1LL * k * x;
      lz += x, mx1 += x;
      if (mx2 > -N) mx2 += x;
    }
    void change(int x) {
      if (!k) return;
      sum += 1LL * x * ted;
      lz_mx += x, mx1 += x;
    }
  } seg[N << 2];
  Node merge(Node a, Node b) {
    if (b.k && a.mx1 < b.mx1) swap(a, b);
    if (a.mx1 == b.mx1) {
      a.ted += b.ted;
      a.mx2 = max(a.mx2, b.mx2);
    } else
      a.mx2 = max(a.mx2, b.mx1);
    a.lz = a.lz_mx = 0;
    a.sum += b.sum;
    a.k += b.k;
    return a;
  }
  void shift(int id) {
    for (auto p : {id << 1, id << 1 | 1}) {
      seg[p].add(seg[id].lz);
      if (seg[p].mx1 + seg[id].lz_mx == seg[id].mx1)
        seg[p].change(seg[id].lz_mx);
    }
    seg[id].lz = seg[id].lz_mx = 0;
  }
  void change(int p, int x, int id = 1, int st = 0, int en = n) {
    if (en - st == 1) {
      seg[id].mx1 = seg[id].sum = x;
      seg[id].ted = seg[id].k = 1;
      return;
    }
    shift(id);
    int mid = st + en >> 1;
    if (p < mid)
      change(p, x, id << 1, st, mid);
    else
      change(p, x, id << 1 | 1, mid, en);
    seg[id] = merge(seg[id << 1], seg[id << 1 | 1]);
  }
  void update(int l, int r, int x, int id = 1, int st = 0, int en = n) {
    if (r <= st || en <= l || seg[id].mx1 <= x) return;
    if (l <= st && en <= r && seg[id].mx2 < x)
      return seg[id].change(x - seg[id].mx1);
    shift(id);
    int mid = st + en >> 1;
    update(l, r, x, id << 1, st, mid);
    update(l, r, x, id << 1 | 1, mid, en);
    seg[id] = merge(seg[id << 1], seg[id << 1 | 1]);
  }
  void add(int l, int r, int x, int id = 1, int st = 0, int en = n) {
    if (r <= st || en <= l) return;
    if (l <= st && en <= r) return seg[id].add(x);
    shift(id);
    int mid = st + en >> 1;
    add(l, r, x, id << 1, st, mid);
    add(l, r, x, id << 1 | 1, mid, en);
    seg[id] = merge(seg[id << 1], seg[id << 1 | 1]);
  }
} L, R;
void add(int p) {
  for (p++; p < N; p += p & -p) fen[p]++;
}
int get(int p) {
  int ans = 0;
  for (p++; p; p -= p & -p) ans += fen[p];
  return ans;
}
void read_input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    p[--x] = i;
  }
}
void solve() {
  for (int i = 0; i < n; i++) {
    R.add(p[i], n, 1), R.update(0, p[i], get(p[i]));
    L.add(p[i], n, -1), L.update(p[i], n, -get(p[i]));
    L.change(p[i], 1), R.change(p[i], i + 1), add(p[i]);
    cout << L.seg[1].sum + R.seg[1].sum - i - 1 << endl;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), solve();
  return 0;
}
