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, a[800000], pos[800000], bit[800000], mn[800000], mn2[800000],
    num[800000], tag[800000];
long long ans, out[800000];
void add(int x) {
  for (; x <= n; x += x & (-x)) ++bit[x];
}
int qry(int x) {
  int ret = 0;
  for (; x; x -= x & (-x)) ret += bit[x];
  return ret;
}
void upd(int x) {
  mn[x] = min(mn[(x << 1)], mn[(x << 1 | 1)]);
  num[x] = 0;
  if (mn[(x << 1)] == mn[x]) num[x] += num[(x << 1)];
  if (mn[(x << 1 | 1)] == mn[x]) num[x] += num[(x << 1 | 1)];
  mn2[x] = 0x3f3f3f3f;
  if (mn[(x << 1)] != mn[x]) mn2[x] = mn[(x << 1)];
  if (mn[(x << 1 | 1)] != mn[x]) mn2[x] = mn[(x << 1 | 1)];
  mn2[x] = min(mn2[x], min(mn2[(x << 1)], mn2[(x << 1 | 1)]));
}
void psd(int x) {
  if (~tag[x]) {
    if (tag[x] > mn[(x << 1)]) {
      mn[(x << 1)] = tag[x];
      tag[(x << 1)] = tag[x];
    }
    if (tag[x] > mn[(x << 1 | 1)]) {
      mn[(x << 1 | 1)] = tag[x];
      tag[(x << 1 | 1)] = tag[x];
    }
    tag[x] = -1;
  }
}
void open(int x, int l, int r, int k) {
  if (l == r) {
    mn[x] = 0;
    num[x] = 1;
    ans += qry(k);
    return;
  }
  int mid = (l + r) >> 1;
  psd(x);
  if (k <= mid) open((x << 1), l, mid, k);
  if (k > mid) open((x << 1 | 1), mid + 1, r, k);
  upd(x);
}
void mdf(int x, int l, int r, int L, int R, int k) {
  if (L > R) return;
  if (mn[x] >= k) return;
  int mid = (l + r) >> 1;
  if (l >= L && r <= R) {
    if (mn2[x] >= k) {
      tag[x] = k;
      ans += (long long)(qry(mn[x]) + 1) * num[x];
      mn[x] = k;
      ans -= (long long)qry(mn[x]) * num[x];
    } else {
      psd(x);
      mdf((x << 1), l, mid, L, R, k);
      mdf((x << 1 | 1), mid + 1, r, L, R, k);
      upd(x);
    }
    return;
  }
  psd(x);
  if (L <= mid) mdf((x << 1), l, mid, L, R, k);
  if (R > mid) mdf((x << 1 | 1), mid + 1, r, L, R, k);
  upd(x);
}
void build(int x, int l, int r) {
  tag[x] = -1;
  if (l == r) {
    mn[x] = mn2[x] = 0x3f3f3f3f;
    num[x] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build((x << 1), l, mid);
  build((x << 1 | 1), mid + 1, r);
  upd(x);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    int j = pos[i];
    add(j);
    open(1, 1, n, j);
    mdf(1, 1, n, j + 1, n, j);
    out[i] += ans;
  }
  memset(bit, 0, sizeof bit);
  build(1, 1, n);
  ans = 0;
  for (int i = 1; i <= n; ++i) {
    int j = n - pos[i] + 1;
    add(j);
    open(1, 1, n, j);
    mdf(1, 1, n, j + 1, n, j);
    out[i] += ans;
  }
  for (int i = 1; i <= n; ++i) printf("%lld\n", out[i] - i);
  return 0;
}
