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;
struct {
  template <typename T>
  inline operator T() {
    T x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) (c == '-') && (f = -1), c = getchar();
    while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f * x;
  }
} gi;
const int N = 2e5 + 5, M = 1e6 + 5;
int c[M], v[M], s[M], tg[M], stg[M];
int le[M], sle[M];
long long sum[M];
void clear(int o, int l, int r) {
  v[o] = s[o] = c[o] = tg[o] = stg[o] = le[o] = sle[o] = sum[o] = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  clear(o << 1, l, mid), clear(o << 1 | 1, mid + 1, r);
}
void psup(int o) {
  c[o] = c[o << 1] + c[o << 1 | 1], le[o] = le[o << 1] + le[o << 1 | 1],
  sum[o] = sum[o << 1] + sum[o << 1 | 1];
  if (v[o << 1] > v[o << 1 | 1]) {
    v[o] = v[o << 1], s[o] = max(s[o << 1], v[o << 1 | 1]);
    le[o] = le[o << 1], sle[o] = sle[o << 1] + le[o << 1 | 1] + sle[o << 1 | 1];
  } else if (v[o << 1] < v[o << 1 | 1]) {
    v[o] = v[o << 1 | 1], s[o] = max(s[o << 1 | 1], v[o << 1]);
    le[o] = le[o << 1 | 1], sle[o] = sle[o << 1 | 1] + le[o << 1] + sle[o << 1];
  } else {
    v[o] = v[o << 1], s[o] = max(s[o << 1], s[o << 1 | 1]);
    le[o] = le[o << 1] + le[o << 1 | 1], sle[o] = sle[o << 1] + sle[o << 1 | 1];
  }
}
void pshn(int o, int t1, int t2, int f) {
  if (!f) t1 = t2;
  v[o] += t1, tg[o] += t1, s[o] += t2, stg[o] += t2;
  sum[o] += 1ll * t1 * le[o] + 1ll * t2 * sle[o];
}
void psdw(int o) {
  int t1 = v[o << 1] >= v[o << 1 | 1], t2 = v[o << 1 | 1] >= v[o << 1];
  pshn(o << 1, tg[o], stg[o], t1);
  pshn(o << 1 | 1, tg[o], stg[o], t2);
  tg[o] = stg[o] = 0;
}
int add(int o, int l, int r, int L, int R) {
  if (l > R || L > r) return 0;
  if (L <= l && R >= r) return pshn(o, 1, 1, 1), c[o];
  psdw(o);
  int mid = (l + r >> 1),
      rs = add(o << 1, l, mid, L, R) + add(o << 1 | 1, mid + 1, r, L, R);
  psup(o);
  return rs;
}
void upd(int o, int l, int r, int p, int x) {
  if (l == r) return sum[o] = v[o] = x, le[o] = c[o] = 1, void();
  psdw(o);
  int mid = (l + r) >> 1;
  if (p <= mid)
    upd(o << 1, l, mid, p, x);
  else
    upd(o << 1 | 1, mid + 1, r, p, x);
  psup(o);
}
void cmin(int o, int l, int r, int L, int R, int va) {
  if (L > r || l > R || va >= v[o]) return;
  if (L <= l && r <= R && va > s[o])
    return pshn(o, min(va - v[o], 0), 0, 1), void();
  psdw(o);
  int mid = (l + r) >> 1;
  cmin(o << 1, l, mid, L, R, va), cmin(o << 1 | 1, mid + 1, r, L, R, va);
  psup(o);
}
int n, a[N], p[N];
long long ans[N];
int main() {
  n = gi;
  for (int i = 1; i <= n; ++i) p[a[i] = gi] = i;
  for (int t = 1; t <= 2; ++t) {
    clear(1, 1, n);
    for (int i = 1, x; i <= n; ++i) {
      x = add(1, 1, n, p[i] + 1, n), upd(1, 1, n, p[i], i + 1);
      cmin(1, 1, n, 1, p[i] - 1, i - x), ans[i] += 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", ans[i] - 1ll * i * (i + 2));
  return 0;
}
