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;
const int N = 150010;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, a[N], pos[N];
struct seg {
  int mx[N << 2], se[N << 2], cnt[N << 2], tag[N << 2], tag_mx[N << 2];
  long long sum[N << 2];
  void up(int x) {
    mx[x] = max(mx[(x << 1)], mx[(x << 1 | 1)]);
    se[x] =
        max(mx[x] == mx[(x << 1)] ? se[(x << 1)] : mx[(x << 1)],
            mx[x] == mx[(x << 1 | 1)] ? se[(x << 1 | 1)] : mx[(x << 1 | 1)]);
    cnt[x] = 0;
    if (mx[x] == mx[(x << 1)]) cnt[x] += cnt[(x << 1)];
    if (mx[x] == mx[(x << 1 | 1)]) cnt[x] += cnt[(x << 1 | 1)];
    sum[x] = sum[(x << 1)] + sum[(x << 1 | 1)];
  }
  void mdf_max(int x, int v) {
    tag_mx[x] += v;
    mx[x] += v;
    sum[x] += 1ll * cnt[x] * v;
  }
  void mdf_add(int x, int l, int r, int v) {
    mx[x] += v;
    se[x] += v;
    tag[x] += v;
    sum[x] += 1ll * (r - l + 1) * v;
  }
  void down(int x, int l, int r) {
    int mid = (l + r) >> 1;
    if (tag[x])
      mdf_add((x << 1), l, mid, tag[x]),
          mdf_add((x << 1 | 1), mid + 1, r, tag[x]), tag[x] = 0;
    if (mx[(x << 1)] + tag_mx[x] == mx[x]) mdf_max((x << 1), tag_mx[x]);
    if (mx[(x << 1 | 1)] + tag_mx[x] == mx[x]) mdf_max((x << 1 | 1), tag_mx[x]);
    tag_mx[x] = 0;
  }
  void modify_min(int L, int R, int v, int x = 1, int l = 1, int r = n) {
    if (L > R || mx[x] <= v) return;
    if (L <= l && r <= R && se[x] < v) return mdf_max(x, v - mx[x]);
    down(x, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) modify_min(L, R, v, (x << 1), l, mid);
    if (R > mid) modify_min(L, R, v, (x << 1 | 1), mid + 1, r);
    up(x);
  }
  void modify_add(int L, int R, int v, int x = 1, int l = 1, int r = n) {
    if (L > R) return;
    if (L <= l && r <= R) return mdf_add(x, l, r, v);
    down(x, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) modify_add(L, R, v, (x << 1), l, mid);
    if (R > mid) modify_add(L, R, v, (x << 1 | 1), mid + 1, r);
    up(x);
  }
  void modify(int p, int v, int x = 1, int l = 1, int r = n) {
    if (l == r) {
      mx[x] = v;
      se[x] = -1e9;
      cnt[x] = 1;
      sum[x] = v;
      return;
    }
    down(x, l, r);
    int mid = (l + r) >> 1;
    p <= mid ? modify(p, v, (x << 1), l, mid)
             : modify(p, v, (x << 1 | 1), mid + 1, r);
    up(x);
  }
  long long query(int p, int x = 1, int l = 1, int r = n) {
    if (l == r) return sum[x];
    down(x, l, r);
    int mid = (l + r) >> 1;
    return p <= mid ? query(p, (x << 1), l, mid)
                    : query(p, (x << 1 | 1), mid + 1, r);
  }
} tl, tr;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = gi(), pos[a[i]] = i;
  for (int i = 1; i <= n; i++) tl.modify(i, n), tr.modify(i, 0);
  for (int i = 1; i <= n; i++) {
    int x = pos[i];
    tr.modify_add(x + 1, n, 1);
    tl.modify_add(x + 1, n, -1);
    int now = tr.query(x);
    tr.modify_min(1, x - 1, now);
    tl.modify_min(x + 1, n, n - now - 1);
    tr.modify(x, i);
    tl.modify(x, n);
    cout << tl.sum[1] + tr.sum[1] - 1ll * n * (n + 1) + i + n - i << '\n';
  }
  return 0;
}
