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;
namespace FGF {
const int N = 2e5 + 5;
int n, a[N], p[N];
struct tree {
  int mx, cm, s, ms;
  long long ta, cta, sum;
} t[N << 2];
long long ans[N];
void build(int ro, int l, int r) {
  t[ro].mx = t[ro].cm = t[ro].s = t[ro].ms = t[ro].ta = t[ro].cta = t[ro].sum =
      0;
  if (l == r) return;
  build(ro << 1, l, ((l + r) >> 1)), build(ro << 1 | 1, ((l + r) >> 1) + 1, r);
}
void tag(int ro, int ta, int ct, int o) {
  if (!o) ta = ct;
  t[ro].mx += ta, t[ro].cm += ct, t[ro].ta += ta, t[ro].cta += ct;
  t[ro].sum += 1ll * ta * t[ro].ms + 1ll * ct * (t[ro].s - t[ro].ms);
}
void pushdown(int ro) {
  int fl = t[ro << 1].mx >= t[ro << 1 | 1].mx,
      fr = t[ro << 1 | 1].mx >= t[ro << 1].mx;
  tag(ro << 1, t[ro].ta, t[ro].cta, fl),
      tag(ro << 1 | 1, t[ro].ta, t[ro].cta, fr);
  t[ro].ta = t[ro].cta = 0;
}
void pushup(int ro) {
  t[ro].s = t[ro << 1].s + t[ro << 1 | 1].s,
  t[ro].sum = t[ro << 1].sum + t[ro << 1 | 1].sum;
  if (t[ro << 1].mx > t[ro << 1 | 1].mx)
    t[ro].mx = t[ro << 1].mx, t[ro].cm = max(t[ro << 1].cm, t[ro << 1 | 1].mx),
    t[ro].ms = t[ro << 1].ms;
  else if (t[ro << 1].mx < t[ro << 1 | 1].mx)
    t[ro].mx = t[ro << 1 | 1].mx,
    t[ro].cm = max(t[ro << 1 | 1].cm, t[ro << 1].mx),
    t[ro].ms = t[ro << 1 | 1].ms;
  else
    t[ro].mx = t[ro << 1].mx, t[ro].cm = max(t[ro << 1 | 1].cm, t[ro << 1].cm),
    t[ro].ms = t[ro << 1 | 1].ms + t[ro << 1].ms;
}
void updat(int ro, int l, int r, int L, int R) {
  if (L <= l && r <= R) return tag(ro, 1, 1, 1);
  pushdown(ro);
  if (L <= ((l + r) >> 1)) updat(ro << 1, l, ((l + r) >> 1), L, R);
  if (R > ((l + r) >> 1)) updat(ro << 1 | 1, ((l + r) >> 1) + 1, r, L, R);
  pushup(ro);
}
void inser(int ro, int l, int r, int x, int y) {
  if (l == r) {
    t[ro].mx = t[ro].sum = y, t[ro].ms = t[ro].s = 1;
    return;
  }
  pushdown(ro);
  x <= ((l + r) >> 1) ? inser(ro << 1, l, ((l + r) >> 1), x, y)
                      : inser(ro << 1 | 1, ((l + r) >> 1) + 1, r, x, y);
  pushup(ro);
}
void getmin(int ro, int l, int r, int L, int R, int y) {
  if (y >= t[ro].mx) return;
  if (L <= l && r <= R && y > t[ro].cm) return tag(ro, y - t[ro].mx, 0, 1);
  pushdown(ro);
  if (L <= ((l + r) >> 1)) getmin(ro << 1, l, ((l + r) >> 1), L, R, y);
  if (R > ((l + r) >> 1)) getmin(ro << 1 | 1, ((l + r) >> 1) + 1, r, L, R, y);
  pushup(ro);
}
int query(int ro, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[ro].s;
  pushdown(ro);
  return (L <= ((l + r) >> 1) ? query(ro << 1, l, ((l + r) >> 1), L, R) : 0) +
         (R > ((l + r) >> 1) ? query(ro << 1 | 1, ((l + r) >> 1) + 1, r, L, R)
                             : 0);
}
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (int k = 1; k <= 2; k++) {
    build(1, 1, n);
    for (int i = 1, x; i <= n; i++) {
      if (p[i] < n) updat(1, 1, n, p[i] + 1, n);
      inser(1, 1, n, p[i], i + 1);
      if (p[i] > 1)
        getmin(1, 1, n, 1, p[i] - 1, x = (query(1, 1, n, 1, p[i] - 1) + 1));
      ans[i] += t[1].sum;
    }
    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));
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
