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;
struct Node {
  long long mx, cmx, mx2, tg, tg2, s;
  int c;
  inline void ladd(long long t) {
    s += t * c;
    tg += t;
    mx += t;
    mx2 += t;
  }
  inline void ladd2(long long t) {
    mx += t;
    s += cmx * t;
    tg2 += t;
  }
} a[150500 << 2];
int n;
inline void up(int u) {
  int l = u << 1, r = u << 1 | 1;
  a[u].s = a[l].s + a[r].s;
  a[u].c = a[l].c + a[r].c;
  if (a[l].mx == a[r].mx) {
    a[u].mx = a[l].mx;
    a[u].cmx = a[l].cmx + a[r].cmx;
    a[u].mx2 = max(a[l].mx2, a[r].mx2);
  } else {
    if (a[l].mx > a[r].mx) swap(l, r);
    a[u].mx = a[r].mx;
    a[u].cmx = a[r].cmx;
    a[u].mx2 = max(a[l].mx, a[r].mx2);
  }
}
void build(int l = 1, int r = n, int u = 1) {
  a[u].c = a[u].tg = a[u].tg2 = 0;
  a[u].s = a[u].mx = 0;
  a[u].mx2 = -1ll << 50;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, u << 1);
  build(mid + 1, r, u << 1 | 1);
}
void ladd(int u) {
  if (a[u].tg2) {
    int l = u << 1, r = u << 1 | 1, mx = max(a[l].mx, a[r].mx);
    if (a[l].mx == mx) a[l].ladd2(a[u].tg2);
    if (a[r].mx == mx) a[r].ladd2(a[u].tg2);
    a[u].tg2 = 0;
  }
  if (a[u].tg) {
    a[u << 1].ladd(a[u].tg);
    a[u << 1 | 1].ladd(a[u].tg);
    a[u].tg = 0;
  }
}
int wfl, wfr, wfc;
void add(int l = 1, int r = n, int u = 1) {
  if (wfl <= l && r <= wfr) {
    a[u].ladd(wfc);
    return;
  }
  int mid = (l + r) >> 1;
  ladd(u);
  if (wfl <= mid) add(l, mid, u << 1);
  if (mid < wfr) add(mid + 1, r, u << 1 | 1);
  up(u);
}
void chg(int l = 1, int r = n, int u = 1) {
  if (wfc >= a[u].mx) return;
  if (wfl <= l && r <= wfr && wfc > a[u].mx2) {
    a[u].ladd2(wfc - a[u].mx);
    return;
  }
  int mid = (l + r) >> 1;
  ladd(u);
  if (wfl <= mid) chg(l, mid, u << 1);
  if (mid < wfr) chg(mid + 1, r, u << 1 | 1);
  up(u);
}
int to;
void cre(int l = 1, int r = n, int u = 1) {
  if (l == r) {
    a[u].mx = a[u].s = wfc;
    a[u].c = a[u].cmx = 1;
    return;
  }
  int mid = (l + r) >> 1;
  ladd(u);
  if (to <= mid)
    cre(l, mid, u << 1);
  else
    cre(mid + 1, r, u << 1 | 1);
  up(u);
}
int qryc(int l = 1, int r = n, int u = 1) {
  if (wfl <= l && r <= wfr) return a[u].c;
  int mid = (l + r) >> 1, ret = 0;
  if (wfl <= mid) ret = qryc(l, mid, u << 1);
  if (mid < wfr) ret += qryc(mid + 1, r, u << 1 | 1);
  return ret;
}
int x[150500], p[150500];
long long ans[150500];
bool cmp(const int &A, const int &B) { return x[A] < x[B]; }
void calc() {
  for (int i = 1; i <= n; i++) p[i] = i;
  sort(p + 1, p + n + 1, cmp);
  build();
  for (int i = 1; i <= n; i++) {
    wfl = 1;
    wfr = p[i] - 1;
    if (wfl <= wfr) {
      wfc = qryc() + 1;
      chg();
    }
    wfl = p[i] + 1;
    wfr = n;
    wfc = 1;
    add();
    to = p[i];
    wfc = i + 1;
    cre();
    ans[i] += a[1].s;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  calc();
  reverse(x + 1, x + n + 1);
  calc();
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] - 1ll * i * (i + 2));
  return 0;
}
