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;
void read(long long &x) {
  char ch = getchar();
  x = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
}
const long long N = 2e5 + 5, M = N << 2;
struct tree {
  long long c[M], v[M], s[M], tg[M], tag[M];
  long long cnt[M], sum[M];
  void clear(long long p, long long l, long long r) {
    c[p] = v[p] = s[p] = tg[p] = 0;
    cnt[p] = sum[p] = tag[p] = 0;
    if (l == r) return;
    long long mid(l + r >> 1);
    clear((p << 1), l, mid), clear((p << 1 | 1), mid + 1, r);
  }
  void pushup(long long p) {
    long long l = (p << 1), r = (p << 1 | 1);
    c[p] = c[l] + c[r], sum[p] = sum[l] + sum[r];
    if (v[l] > v[r]) {
      v[p] = v[l], cnt[p] = cnt[l];
      s[p] = max(s[l], v[r]);
    } else if (v[l] < v[r]) {
      v[p] = v[r], cnt[p] = cnt[r];
      s[p] = max(v[l], s[r]);
    } else {
      v[p] = v[l], cnt[p] = cnt[l] + cnt[r];
      s[p] = max(s[l], s[r]);
    }
  }
  void make(long long p, long long v1, long long v2, long long o) {
    if (!o) v1 = v2;
    v[p] += v1, tg[p] += v1, s[p] += v2, tag[p] += v2;
    sum[p] += v1 * cnt[p] + v2 * (c[p] - cnt[p]);
  }
  void pushdown(long long p) {
    long long l = (p << 1), r = (p << 1 | 1), o = v[r] >= v[l];
    make(l, tg[p], tag[p], v[l] >= v[r]);
    make(r, tg[p], tag[p], o);
    tg[p] = tag[p] = 0;
  }
  long long add(long long p, long long l, long long r, long long ql,
                long long qr) {
    if (ql <= l && qr >= r) return make(p, 1, 1, 1), c[p];
    long long mid(l + r >> 1), e = 0;
    pushdown(p);
    if (ql <= mid) e += add((p << 1), l, mid, ql, qr);
    if (qr > mid) e += add((p << 1 | 1), mid + 1, r, ql, qr);
    return pushup(p), e;
  }
  void modify(long long p, long long l, long long r, long long x,
              long long val) {
    if (l == r) {
      sum[p] = v[p] = val, cnt[p] = c[p] = 1;
      return;
    }
    long long mid(l + r >> 1);
    pushdown(p);
    if (x <= mid)
      modify((p << 1), l, mid, x, val);
    else
      modify((p << 1 | 1), mid + 1, r, x, val);
    pushup(p);
  }
  void up(long long p, long long l, long long r, long long ql, long long qr,
          long long val) {
    if (val >= v[p]) return;
    if (ql <= l && qr >= r && val > s[p]) {
      make(p, val - v[p], 0, 1);
      return;
    }
    long long mid(l + r >> 1);
    pushdown(p);
    if (ql <= mid) up((p << 1), l, mid, ql, qr, val);
    if (qr > mid) up((p << 1 | 1), mid + 1, r, ql, qr, val);
    pushup(p);
  }
} Y;
long long n, a[N], w[N], res[N];
void solve() {
  Y.clear(1, 1, n);
  for (long long i = 1, x; i <= n; ++i, x = 0) {
    if (w[i] < n) x = Y.add(1, 1, n, w[i] + 1, n);
    Y.modify(1, 1, n, w[i], i + 1);
    if (w[i] > 1) Y.up(1, 1, n, 1, w[i] - 1, i - x);
    res[i] += Y.sum[1];
  }
}
signed main() {
  read(n);
  for (long long i = 1; i <= n; ++i) read(a[i]), w[a[i]] = i;
  solve();
  for (long long i = 1; i <= n; ++i) w[i] = n - w[i] + 1;
  solve();
  for (long long i = 1; i <= n; ++i) printf("%lld\n", res[i] - i * (i + 2));
  return 0;
}
