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 = 150005;
int n;
int p[maxn], pos[maxn];
class Segtree {
 public:
  int mx1[maxn * 4], mx2[maxn * 4], mxcnt[maxn * 4], sumcnt[maxn * 4],
      tag[maxn * 4];
  long long sum[maxn * 4];
  void pushup(int cur) {
    if (mx1[cur << 1] > mx1[cur << 1 | 1]) {
      mx1[cur] = mx1[cur << 1];
      mxcnt[cur] = mxcnt[cur << 1];
      mx2[cur] = max(mx2[cur << 1], mx1[cur << 1 | 1]);
    } else if (mx1[cur << 1 | 1] > mx1[cur << 1]) {
      mx1[cur] = mx1[cur << 1 | 1];
      mxcnt[cur] = mxcnt[cur << 1 | 1];
      mx2[cur] = max(mx2[cur << 1 | 1], mx1[cur << 1]);
    } else {
      mx1[cur] = mx1[cur << 1];
      mxcnt[cur] = mxcnt[cur << 1] + mxcnt[cur << 1 | 1];
      mx2[cur] = max(mx2[cur << 1], mx2[cur << 1 | 1]);
    }
    sum[cur] = sum[cur << 1] + sum[cur << 1 | 1];
    sumcnt[cur] = sumcnt[cur << 1] + sumcnt[cur << 1 | 1];
  }
  void apply_mx(int cur, int val) {
    sum[cur] -= 1LL * mxcnt[cur] * (mx1[cur] - val);
    mx1[cur] = val;
  }
  void apply_sum(int cur, int val) {
    sum[cur] += 1LL * sumcnt[cur] * val;
    if (mx1[cur]) mx1[cur] += val;
    if (mx2[cur]) mx2[cur] += val;
    tag[cur] += val;
  }
  void pushdown(int cur) {
    if (tag[cur]) {
      apply_sum(cur << 1, tag[cur]);
      apply_sum(cur << 1 | 1, tag[cur]);
      tag[cur] = 0;
    }
    if (mx1[cur] < mx1[cur << 1] && mx1[cur] >= mx2[cur << 1])
      apply_mx(cur << 1, mx1[cur]);
    if (mx1[cur] < mx1[cur << 1 | 1] && mx1[cur] >= mx2[cur << 1 | 1])
      apply_mx(cur << 1 | 1, mx1[cur]);
  }
  void build(int cur, int l, int r) {
    if (l == r) {
      mxcnt[cur] = 1;
      return;
    }
    int mid = l + r >> 1;
    build(cur << 1, l, mid);
    build(cur << 1 | 1, mid + 1, r);
    pushup(cur);
  }
  void update_mx(int cur, int l, int r, int vl, int vr, int val) {
    if (vl > vr) return;
    if (mx1[cur] <= val) return;
    if (l >= vl && r <= vr) {
      if (mx2[cur] <= val) {
        apply_mx(cur, val);
        return;
      }
    }
    pushdown(cur);
    int mid = l + r >> 1;
    if (mid >= vl) update_mx(cur << 1, l, mid, vl, vr, val);
    if (mid < vr) update_mx(cur << 1 | 1, mid + 1, r, vl, vr, val);
    pushup(cur);
  }
  void update_sum(int cur, int l, int r, int vl, int vr, int val) {
    if (vl > vr) return;
    if (l >= vl && r <= vr) {
      apply_sum(cur, val);
      return;
    }
    pushdown(cur);
    int mid = l + r >> 1;
    if (mid >= vl) update_sum(cur << 1, l, mid, vl, vr, val);
    if (mid < vr) update_sum(cur << 1 | 1, mid + 1, r, vl, vr, val);
    pushup(cur);
  }
  void modify(int cur, int l, int r, int k, int val) {
    if (l == r) {
      mx1[cur] = sum[cur] = val;
      mx2[cur] = 0;
      mxcnt[cur] = sumcnt[cur] = 1;
      return;
    }
    pushdown(cur);
    int mid = l + r >> 1;
    if (k <= mid)
      modify(cur << 1, l, mid, k, val);
    else
      modify(cur << 1 | 1, mid + 1, r, k, val);
    pushup(cur);
  }
  int query(int cur, int l, int r, int vl, int vr) {
    if (vl > vr) return 0;
    if (l >= vl && r <= vr) return sumcnt[cur];
    pushdown(cur);
    int mid = l + r >> 1;
    int sl = mid >= vl ? query(cur << 1, l, mid, vl, vr) : 0;
    int sr = mid < vr ? query(cur << 1 | 1, mid + 1, r, vl, vr) : 0;
    return sl + sr;
  }
} Mx, Mn;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    pos[p[i]] = i;
  }
  Mx.build(1, 1, n);
  Mn.build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int rk = Mx.query(1, 1, n, 1, pos[i] - 1) + 1;
    Mx.update_mx(1, 1, n, 1, pos[i] - 1, rk - 1);
    Mn.update_mx(1, 1, n, pos[i] + 1, n, n + 1 - rk);
    Mx.update_sum(1, 1, n, pos[i] + 1, n, 1);
    Mn.update_sum(1, 1, n, pos[i] + 1, n, -1);
    Mx.modify(1, 1, n, pos[i], i);
    Mn.modify(1, 1, n, pos[i], n + 1 - 1);
    printf("%lld\n", Mx.sum[1] + Mn.sum[1] - 1LL * i * n);
  }
  return 0;
}
