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;
int n, a[150005], b[150005];
long long ans[150005], sm[150005 << 2], ad[150005 << 2], adm[150005 << 2];
int mx[150005 << 2], cmx[150005 << 2], hdm[150005 << 2], hd[150005 << 2];
void Build(int u, int l, int r) {
  sm[u] = ad[u] = adm[u] = mx[u] = cmx[u] = hdm[u] = hd[u] = 0;
  if (l == r) return;
  int m = l + r >> 1;
  Build(u << 1, l, m), Build(u << 1 | 1, m + 1, r);
}
void dtmp(int u, int v) {
  adm[u] += v, sm[u] += v * 1ll * hdm[u];
  if (hd[u]) mx[u] += v;
}
void dtp(int u, int v) {
  ad[u] += v, sm[u] += v * 1ll * hd[u];
  if (hd[u]) mx[u] += v;
  if (hd[u] && hd[u] != hdm[u]) cmx[u] += v;
}
void dt(int u) {
  if (ad[u]) {
    dtp(u << 1, ad[u]), dtp(u << 1 | 1, ad[u]);
    ad[u] = 0;
  }
  if (adm[u]) {
    if (mx[u] == mx[u << 1] + adm[u]) dtmp(u << 1, adm[u]);
    if (mx[u] == mx[u << 1 | 1] + adm[u]) dtmp(u << 1 | 1, adm[u]);
    adm[u] = 0;
  }
}
void upd(int u) {
  sm[u] = sm[u << 1] + sm[u << 1 | 1];
  mx[u] = max(mx[u << 1], mx[u << 1 | 1]);
  hdm[u] = cmx[u] = 0;
  if (mx[u] == mx[u << 1])
    hdm[u] += hdm[u << 1], cmx[u] = max(cmx[u], cmx[u << 1]);
  else
    cmx[u] = max(cmx[u], mx[u << 1]);
  if (mx[u] == mx[u << 1 | 1])
    hdm[u] += hdm[u << 1 | 1], cmx[u] = max(cmx[u], cmx[u << 1 | 1]);
  else
    cmx[u] = max(cmx[u], mx[u << 1 | 1]);
}
void ckm(int u, int l, int r, int ql, int qr, int v) {
  if (l > qr || ql > r || ql > qr || v >= mx[u]) return;
  if (ql <= l && r <= qr && v > cmx[u]) return (void)(dtmp(u, v - mx[u]));
  int m = l + r >> 1;
  dt(u);
  ckm(u << 1, l, m, ql, qr, v), ckm(u << 1 | 1, m + 1, r, ql, qr, v);
  upd(u);
}
void ins(int u, int l, int r, int p, int v) {
  if (l == r) return (void)(sm[u] = mx[u] = v, hd[u] = hdm[u] = 1);
  int m = l + r >> 1;
  dt(u);
  p <= m ? ins(u << 1, l, m, p, v) : ins(u << 1 | 1, m + 1, r, p, v);
  upd(u);
  hd[u] = hd[u << 1] + hd[u << 1 | 1];
}
void add(int u, int l, int r, int ql, int qr, int v) {
  if (l > qr || ql > r || ql > qr) return;
  if (ql <= l && r <= qr) return (void)(dtp(u, v));
  int m = l + r >> 1;
  dt(u);
  add(u << 1, l, m, ql, qr, v), add(u << 1 | 1, m + 1, r, ql, qr, v);
  upd(u);
}
int qry(int u, int l, int r, int ql, int qr) {
  if (l > qr || ql > r) return 0;
  if (ql <= l && r <= qr) return hd[u];
  int m = l + r >> 1;
  return qry(u << 1, l, m, ql, qr) + qry(u << 1 | 1, m + 1, r, ql, qr);
}
void Solve() {
  Build(1, 1, n);
  for (int i = (1), LIM = (n); i <= LIM; i++) {
    int p = qry(1, 1, n, 1, b[i] - 1) + 1;
    ckm(1, 1, n, 1, b[i] - 1, p - 1);
    ins(1, 1, n, b[i], i);
    add(1, 1, n, b[i] + 1, n, 1);
    ans[i] += sm[1];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1), LIM = (n); i <= LIM; i++) scanf("%d", &a[i]), b[a[i]] = i;
  Solve();
  reverse(a + 1, a + n + 1);
  for (int i = (1), LIM = (n); i <= LIM; i++) b[a[i]] = i;
  for (int i = (1), LIM = (n); i <= LIM; i++) ans[i] -= 1ll * i * i;
  Solve();
  for (int i = (1), LIM = (n); i <= LIM; i++) printf("%lld\n", ans[i]);
}
