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 = 200005;
int n, a[N], p[N];
long long ans;
struct nd {
  int m0, m1, c;
};
inline nd operator+(nd a, nd b) {
  if (a.m0 < b.m0) swap(a, b);
  if (a.m0 == b.m0)
    return {a.m0, max(a.m1, b.m1), a.c + b.c};
  else
    return {a.m0, max(a.m1, b.m0), a.c};
};
struct seg {
  int b[N];
  void add(int x, int v) {
    for (int i = x; i <= n; i += i & (-i)) b[i] += v;
  }
  int sum(int x) {
    int v = 0;
    for (int i = x; i; i -= i & (-i)) v += b[i];
    return v;
  }
  nd tr[N << 2];
  int tag[N << 2];
  void pt(int x, int y) {
    if (tr[x].m0 > tr[y].m0) tr[x].m0 = tr[y].m0;
  }
  void pd(int k) {
    if (tag[k] > tr[k].m0) {
      pt(k << 1, k);
      pt(k << 1 | 1, k);
      tag[k] = tr[k].m0;
    }
  }
  void pu(int k) {
    tr[k] = tr[k << 1] + tr[k << 1 | 1];
    tag[k] = tr[k].m0;
  }
  void build(int k, int l, int r) {
    tag[k] = 1e9;
    if (l == r) return;
    int mid = l + r >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
  }
  void upd(int k, int l, int r, int a, int b, int v) {
    int mid = l + r >> 1;
    if (l == a && r == b) {
      if (v >= tr[k].m0) return;
      if (v < tr[k].m1) {
        pd(k);
        upd(k << 1, l, mid, a, mid, v);
        upd(k << 1 | 1, mid + 1, r, mid + 1, b, v);
        pu(k);
      } else {
        ans -= 1ll * tr[k].c * (sum(tr[k].m0) - sum(v));
        tr[k].m0 = v;
      }
      return;
    }
    pd(k);
    if (b <= mid)
      upd(k << 1, l, mid, a, b, v);
    else if (a > mid)
      upd(k << 1 | 1, mid + 1, r, a, b, v);
    else
      upd(k << 1, l, mid, a, mid, v),
          upd(k << 1 | 1, mid + 1, r, mid + 1, b, v);
    pu(k);
  }
  void ch(int k, int l, int r, int x) {
    if (l == r) {
      tr[k] = {n, 0, 1};
      return;
    }
    pd(k);
    int mid = l + r >> 1;
    if (x <= mid)
      ch(k << 1, l, mid, x);
    else
      ch(k << 1 | 1, mid + 1, r, x);
    pu(k);
  }
} s1, s2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  s1.build(1, 1, n);
  s2.build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int x = p[i];
    if (x > 1) s1.upd(1, 1, n, 1, x - 1, x - 1);
    if (x < n) s2.upd(1, 1, n, 1, n - x, n - x);
    s1.add(x, 1);
    s2.add(n - x + 1, 1);
    s1.ch(1, 1, n, x);
    s2.ch(1, 1, n, n - x + 1);
    ans += i;
    printf("%lld\n", ans);
  }
  return 0;
}
