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 maxn = 1.5e5 + 5;
const int inf = 1e9 + 7;
int n;
int p[maxn];
long long ans[maxn];
int mx[maxn << 2], smx[maxn << 2], cnt[maxn << 2];
int bit[maxn];
long long tans = 0;
inline int lowbit(int x) { return x & (-x); }
void add(int pos, int v) {
  for (int i = pos; i <= n; i += lowbit(i)) bit[i] += v;
}
int get(int pos) {
  int tv = 0;
  for (int i = pos; i; i -= lowbit(i)) tv += bit[i];
  return tv;
}
void updata(int rt) {
  if (mx[rt << 1 | 1] == mx[rt << 1]) {
    mx[rt] = mx[rt << 1];
    cnt[rt] = cnt[rt << 1] + cnt[rt << 1 | 1];
    smx[rt] = max(smx[rt << 1], smx[rt << 1 | 1]);
  } else if (mx[rt << 1] > mx[rt << 1 | 1]) {
    mx[rt] = mx[rt << 1];
    cnt[rt] = cnt[rt << 1];
    smx[rt] = max(smx[rt << 1], mx[rt << 1 | 1]);
  } else {
    mx[rt] = mx[rt << 1 | 1];
    cnt[rt] = cnt[rt << 1 | 1];
    smx[rt] = max(mx[rt << 1], smx[rt << 1 | 1]);
  }
}
void color(int rt, int v) {
  if (mx[rt] <= v) return;
  mx[rt] = v;
}
void pushcol(int rt) {
  color(rt << 1, mx[rt]);
  color(rt << 1 | 1, mx[rt]);
}
void build(int l, int r, int rt) {
  if (l == r) {
    cnt[rt] = 1;
    mx[rt] = 0;
    smx[rt] = -inf;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  updata(rt);
}
void modifyPos(int l, int r, int rt, int pos, int v) {
  if (l == r) {
    if (v) {
      tans += get(v) - get(l - 1);
    } else if (mx[rt]) {
      tans -= get(mx[rt]) - get(l - 1);
    }
    mx[rt] = v;
    smx[rt] = -inf;
    cnt[rt] = 1;
    return;
  }
  pushcol(rt);
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modifyPos(l, mid, rt << 1, pos, v);
  else
    modifyPos(mid + 1, r, rt << 1 | 1, pos, v);
  updata(rt);
}
void modifyLess(int l, int r, int rt, int nl, int nr, int v) {
  if (nl > nr || mx[rt] <= v) return;
  if (nl <= l && nr >= r && smx[rt] < v) {
    tans += (long long)cnt[rt] * (get(v) - get(mx[rt]));
    color(rt, v);
    return;
  }
  pushcol(rt);
  int mid = (l + r) >> 1;
  if (nl <= mid) modifyLess(l, mid, rt << 1, nl, nr, v);
  if (nr > mid) modifyLess(mid + 1, r, rt << 1 | 1, nl, nr, v);
  updata(rt);
}
void solve() {
  memset(bit, 0, sizeof(bit));
  build(1, n, 1);
  tans = 0;
  for (int i = 1; i <= n; i++) {
    modifyPos(1, n, 1, p[i], 0);
    if (p[i] > 1) modifyLess(1, n, 1, 1, p[i] - 1, p[i] - 1);
    if (p[i] < n) modifyPos(1, n, 1, p[i] + 1, n);
    add(p[i], 1);
    ans[i] += tans;
  }
}
int a[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), p[a[i]] = i;
  solve();
  for (int i = 1; i <= n; i++) p[i] = n - p[i] + 1;
  solve();
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] + i);
}
