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 read();
int n, a[200005], p[200005];
struct Seg {
  int c[1000006], v[1000006], s[1000006], tg[1000006], stg[1000006];
  int len[1000006], slen[1000006];
  long long sum[1000006];
  void clear(int l, int r, int k) {
    v[k] = s[k] = c[k] = tg[k] = stg[k] = len[k] = slen[k] = sum[k] = 0;
    if (l == r) return;
    int mid = l + r >> 1;
    clear(l, mid, k << 1), clear(mid + 1, r, k << 1 | 1);
  }
  void update(int k) {
    int L = k << 1, R = k << 1 | 1;
    c[k] = c[L] + c[R], len[k] = len[L] + len[R], sum[k] = sum[L] + sum[R];
    if (v[L] > v[R]) {
      v[k] = v[L], s[k] = max(s[L], v[R]);
      len[k] = len[L], slen[k] = slen[L] + len[R] + slen[R];
    } else if (v[L] < v[R]) {
      v[k] = v[R], s[k] = max(v[L], s[R]);
      len[k] = len[R], slen[k] = slen[R] + len[L] + slen[L];
    } else {
      v[k] = v[L], s[k] = max(s[L], s[R]);
      len[k] = len[L] + len[R], slen[k] = slen[R] + slen[L];
    }
  }
  void Tag(int k, int t1, int t2, int ctm) {
    if (!ctm) t1 = t2;
    v[k] += t1, tg[k] += t1, s[k] += t2, stg[k] += t2;
    sum[k] += 1ll * t1 * len[k] + 1ll * t2 * slen[k];
  }
  void psd(int k) {
    int L = k << 1, R = k << 1 | 1, mr = v[R] >= v[L];
    Tag(L, tg[k], stg[k], v[L] >= v[R]);
    Tag(R, tg[k], stg[k], mr), tg[k] = stg[k] = 0;
  }
  int add(int l, int r, int k, int st, int en) {
    if (st > r || en < l) return 0;
    if (st <= l && en >= r) return Tag(k, 1, 1, 1), c[k];
    int mid = (psd(k), l + r >> 1), rt = 0;
    rt = add(l, mid, k << 1, st, en) + add(mid + 1, r, k << 1 | 1, st, en),
    update(k);
    return rt;
  }
  void modi(int l, int r, int k, int p, int tv) {
    if (l == r) return sum[k] = v[k] = tv, len[k] = c[k] = 1, void();
    int mid = (psd(k), l + r >> 1);
    p <= mid ? modi(l, mid, k << 1, p, tv)
             : modi(mid + 1, r, k << 1 | 1, p, tv),
        update(k);
  }
  void pmin(int l, int r, int k, int st, int en, int tv) {
    if (st > r || en < l || tv >= v[k]) return;
    if (st <= l && en >= r && tv > s[k])
      return Tag(k, min(tv - v[k], 0), 0, 1), void();
    int mid = (psd(k), l + r >> 1);
    pmin(l, mid, k << 1, st, en, tv), pmin(mid + 1, r, k << 1 | 1, st, en, tv),
        update(k);
  }
  void prt(int l, int r, int k) {
    if (l == r) return c[k] ? printf("%d ", v[k]) : 0, void();
    int mid = (psd(k), l + r >> 1);
    prt(l, mid, k << 1), prt(mid + 1, r, k << 1 | 1);
  }
} seg;
long long res[200005];
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) p[a[i] = read()] = i;
  for (int t = 1; t <= 2; ++t) {
    seg.clear(1, n, 1);
    for (int i = 1, x; i <= n; ++i) {
      x = seg.add(1, n, 1, p[i] + 1, n), seg.modi(1, n, 1, p[i], i + 1);
      seg.pmin(1, n, 1, 1, p[i] - 1, i - x), res[i] += seg.sum[1];
    }
    for (int i = 1; i <= n; ++i) p[i] = n - p[i] + 1;
  }
  for (int i = 1; i <= n; ++i) printf("%lld\n", res[i] - 1ll * i * (i + 2));
  return 0;
}
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') f = (c == '-') ? -1 : f, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
