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 a[200010], p[200010];
int n;
struct Segment_tree {
  long long sum[800010];
  int lazy[800010];
  long long tag[800010];
  int min1[800010], min2[800010], num[800010];
  void Build(int now, int l, int r, int val) {
    tag[now] = 0;
    if (l == r) {
      min1[now] = val, num[now] = 1, min2[now] = 0x3f3f3f3f, sum[now] = val;
      return;
    }
    int mid = l + r >> 1;
    Build(now << 1, l, mid, val), Build(now << 1 | 1, mid + 1, r, val);
    pushup(now);
  }
  void pushdown(int now, int l, int r) {
    int val = min1[now], mid = l + r >> 1;
    min1[now << 1] += tag[now], min2[now << 1] += tag[now],
        tag[now << 1] += tag[now];
    sum[now << 1] += 1ll * (mid - l + 1) * tag[now];
    if (min1[now << 1] < val) {
      sum[now << 1] += 1ll * (val - min1[now << 1]) * num[now << 1];
      min1[now << 1] = val;
    }
    min1[now << 1 | 1] += tag[now], min2[now << 1 | 1] += tag[now],
        tag[now << 1 | 1] += tag[now];
    sum[now << 1 | 1] += 1ll * (r - mid) * tag[now];
    if (min1[now << 1 | 1] < val) {
      sum[now << 1 | 1] += 1ll * (val - min1[now << 1 | 1]) * num[now << 1 | 1];
      min1[now << 1 | 1] = val;
    }
    tag[now] = 0;
  }
  void pushup(int now) {
    sum[now] = sum[now << 1] + sum[now << 1 | 1];
    min1[now] = min1[now << 1], num[now] = num[now << 1],
    min2[now] = min2[now << 1];
    if (min1[now << 1 | 1] < min1[now]) {
      min2[now] = min1[now], min1[now] = min1[now << 1 | 1],
      num[now] = num[now << 1 | 1];
    } else if (min1[now << 1 | 1] == min1[now]) {
      num[now] += num[now << 1 | 1];
    } else
      min2[now] = min(min1[now << 1 | 1], min2[now]);
    min2[now] = min(min2[now], min2[now << 1 | 1]);
  }
  void Update(int now, int l, int r, int L, int R, int x) {
    if (L > R) return;
    if (l == L && r == R) {
      if (x <= min1[now]) return;
      if (x < min2[now]) {
        sum[now] += 1ll * (x - min1[now]) * num[now];
        min1[now] = x;
        return;
      }
      pushdown(now, l, r);
      int mid = l + r >> 1;
      Update(now << 1, l, mid, l, mid, x),
          Update(now << 1 | 1, mid + 1, r, mid + 1, r, x);
      pushup(now);
      return;
    }
    pushdown(now, l, r);
    int mid = l + r >> 1;
    if (R <= mid)
      Update(now << 1, l, mid, L, R, x);
    else if (L > mid)
      Update(now << 1 | 1, mid + 1, r, L, R, x);
    else
      Update(now << 1, l, mid, L, mid, x),
          Update(now << 1 | 1, mid + 1, r, mid + 1, R, x);
    pushup(now);
  }
  void Etadpu(int now, int l, int r, int L, int R, int x) {
    if (L > R) return;
    if (l == L && r == R) {
      tag[now] += x, min1[now] += x, min2[now] += x,
          sum[now] += 1ll * x * (r - l + 1);
      return;
    }
    pushdown(now, l, r);
    int mid = l + r >> 1;
    if (R <= mid)
      Etadpu(now << 1, l, mid, L, R, x);
    else if (L > mid)
      Etadpu(now << 1 | 1, mid + 1, r, L, R, x);
    else
      Etadpu(now << 1, l, mid, L, mid, x),
          Etadpu(now << 1 | 1, mid + 1, r, mid + 1, R, x);
    pushup(now);
  }
  void Modify(int now, int l, int r, int pos, int x) {
    if (l == r) {
      min1[now] = x, num[now] = 1, min2[now] = 0x3f3f3f3f, sum[now] = x;
      return;
    }
    pushdown(now, l, r);
    int mid = l + r >> 1;
    if (pos <= mid)
      Modify(now << 1, l, mid, pos, x);
    else
      Modify(now << 1 | 1, mid + 1, r, pos, x);
    pushup(now);
  }
} segl, segr;
int T[200010];
void Modify(int x, int v) {
  while (x <= n + 5) T[x] += v, x += x & -x;
}
int Query(int x) {
  int res = 0;
  while (x) res += T[x], x -= x & -x;
  return res;
}
int get_num(int l, int r) {
  if (l > r) return 0;
  return r - l + 1 - (Query(r) - Query(l - 1));
}
int get_real(int l, int r) {
  if (l > r) return 0;
  return Query(r) - Query(l - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  segl.Build(1, 1, n, n + 1), segr.Build(1, 1, n, 0);
  long long basel = 0, baser = 0;
  for (int it = 1; it <= n; it++) {
    int pos = p[it], k = Query(pos) + 1;
    segl.Etadpu(1, 1, n, pos + 1, n, 1), basel -= get_num(pos + 1, n);
    segl.Update(1, 1, n, pos + 1, n, k);
    segr.Update(1, 1, n, 1, pos - 1, -k);
    segr.Etadpu(1, 1, n, pos + 1, n, -1), baser += get_num(pos + 1, n);
    segl.Modify(1, 1, n, pos, 0), segr.Modify(1, 1, n, pos, -it - 1);
    basel += get_real(1, pos - 1), baser -= get_real(1, pos - 1);
    long long ans = -segr.sum[1] - segl.sum[1] - basel - baser;
    printf("%lld ", ans - it + 1ll * (n - it) * (n + 1));
    Modify(pos, 1);
  }
  printf("\n");
  return 0;
}
