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 = 150005;
int n, p[N], a[N], b[N];
void change(int x, int v) {
  for (; x <= n; x += x & (-x)) b[x] += v;
}
int ask(int x) {
  int ans = 0;
  for (; x; x -= x & (-x)) ans += b[x];
  return ans;
}
int atg[N * 4], mtg[N * 4], sz[N * 4];
int mn0[N * 4], mn1[N * 4], mns[N * 4];
long long sum[N * 4], ans[N];
void adv(int k, int v) {
  atg[k] += v;
  mtg[k] += v;
  mn0[k] += v;
  mn1[k] += v;
  sum[k] += 1ll * sz[k] * v;
}
void mxv(int k, int v) {
  mtg[k] = max(mtg[k], v);
  if (mn0[k] >= v || !sz[k]) return;
  assert(mn1[k] > v);
  sum[k] += 1ll * mns[k] * (v - mn0[k]);
  mn0[k] = v;
}
void pushdown(int k) {
  if (atg[k]) {
    adv(k * 2, atg[k]);
    adv(k * 2 + 1, atg[k]);
    atg[k] = 0;
  }
  if (mtg[k]) {
    mxv(k * 2, mtg[k]);
    mxv(k * 2 + 1, mtg[k]);
    mtg[k] = 0;
  }
}
void pushup(int k) {
  sum[k] = sum[k * 2] + sum[k * 2 + 1];
  sz[k] = sz[k * 2] + sz[k * 2 + 1];
  mn0[k] = min(mn0[k * 2], mn0[k * 2 + 1]);
  mn1[k] = min(mn0[k * 2] == mn0[k] ? mn1[k * 2] : mn0[k * 2],
               mn0[k * 2 + 1] == mn0[k] ? mn1[k * 2 + 1] : mn0[k * 2 + 1]);
  mns[k] = (mn0[k] == mn0[k * 2] ? mns[k * 2] : 0) +
           (mn0[k] == mn0[k * 2 + 1] ? mns[k * 2 + 1] : 0);
}
void changeP(int k, int l, int r, int p) {
  if (l == r) {
    mns[k] = sz[k] = 1;
    mn0[k] = 0;
    return;
  }
  pushdown(k);
  int mid = (l + r) / 2;
  if (p <= mid)
    changeP(k * 2, l, mid, p);
  else
    changeP(k * 2 + 1, mid + 1, r, p);
  pushup(k);
}
void changev(int k, int l, int r, int x, int y, int v) {
  if (l == x && r == y) return adv(k, v);
  pushdown(k);
  int mid = (l + r) / 2;
  if (y <= mid)
    changev(k * 2, l, mid, x, y, v);
  else if (x > mid)
    changev(k * 2 + 1, mid + 1, r, x, y, v);
  else {
    changev(k * 2, l, mid, x, mid, v);
    changev(k * 2 + 1, mid + 1, r, mid + 1, y, v);
  }
  pushup(k);
}
void changemx(int k, int l, int r, int x, int y, int v) {
  if (mn0[k] >= v || !sz[k]) return;
  if (l == x && r == y && mn1[k] > v) return mxv(k, v);
  pushdown(k);
  int mid = (l + r) / 2;
  if (y <= mid)
    changemx(k * 2, l, mid, x, y, v);
  else if (x > mid)
    changemx(k * 2 + 1, mid + 1, r, x, y, v);
  else {
    changemx(k * 2, l, mid, x, mid, v);
    changemx(k * 2 + 1, mid + 1, r, mid + 1, y, v);
  }
  pushup(k);
}
void solve() {
  memset(sum, 0, sizeof(sum));
  memset(atg, 0, sizeof(atg));
  memset(mtg, 0, sizeof(mtg));
  memset(mn0, 60, sizeof(mn0));
  memset(mn1, 60, sizeof(mn1));
  memset(mns, 0, sizeof(mns));
  memset(sz, 0, sizeof(sz));
  memset(b, 0, sizeof(b));
  for (int i = (int)(1); i <= (int)(n); i++) p[a[i]] = i;
  for (int i = (int)(1); i <= (int)(n); i++) {
    int x = p[i], v = ask(x) + 1;
    changeP(1, 1, n, x);
    if (x != n) {
      changev(1, 1, n, x + 1, n, 1);
      changemx(1, 1, n, x + 1, n, v);
    }
    ans[i] += 1ll * i * (i - 1) / 2 - sum[1];
    change(x, 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  solve();
  reverse(a + 1, a + n + 1);
  solve();
  for (int i = (int)(1); i <= (int)(n); i++) printf("%lld\n", ans[i] + i);
}
