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 mxN = 1.5e5, B = 350, mxBC = (mxN - 1) / B + 1;
int n, a[mxN], bc, mn[mxBC], x[mxN];
long long ans[mxN + 1];
bool b[mxN];
vector<int> u[mxBC];
vector<array<int, 2>> v[mxBC];
void psh(int i) {
  for (int j = i * B; j < min((i + 1) * B, n); ++j)
    if (b[j]) ans[j] += u[i].size();
  for (int j = 0, k = v[i].size(); j < u[i].size() || k;) {
    if (j < u[i].size() && (!k || u[i][j] > v[i][k - 1][0]))
      ans[u[i][j++]] -= v[i].size() - k;
    else
      ans[x[v[i][--k][1]]] -= j;
  }
  u[i].clear();
  while (v[i].size() && v[i].back()[0] > mn[i]) v[i].pop_back();
  for (int j = i * B; j < min((i + 1) * B, n); ++j)
    if (b[j] && x[j] > mn[i]) v[i].push_back({x[j] = mn[i], j});
  mn[i] = n;
}
void solve() {
  fill(x, x + n, n);
  memset(b, 0, n);
  for (int i = 0; i < n; ++i) {
    int d = a[i] / B;
    for (int j = 0; j < d; ++j) mn[j] = min(a[i], mn[j]);
    for (int j = d + 1; j < bc; ++j) u[j].push_back(mn[j]);
    psh(d);
    vector<array<int, 2>> w;
    int k = 0;
    for (; k < v[d].size() && v[d][k][0] < a[i]; ++k) w.push_back(v[d][k]);
    for (int j = d * B; j < a[i]; ++j)
      if (b[j] && x[j] > a[i]) w.push_back({x[j] = a[i], j});
    for (; k < v[d].size(); ++k)
      if (v[d][k][1] > a[i]) w.push_back(v[d][k]);
    v[d] = w;
    for (int j = a[i] + 1; j < min((d + 1) * B, n); ++j) {
      if (b[j]) {
        ++ans[j];
        --ans[x[j]];
      }
    }
    v[d].insert(lower_bound(v[d].begin(), v[d].end(), array<int, 2>{n, a[i]}),
                {n, a[i]});
    b[a[i]] = 1;
  }
  for (int i = 0; i < bc; ++i) {
    psh(i);
    v[i].clear();
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  bc = (n - 1) / B + 1;
  for (int i = 0; i < n; ++i) cin >> a[i], --a[i];
  solve();
  reverse(a, a + n);
  solve();
  for (int i = 0; i < n; ++i) {
    ans[i + 1] += ans[i];
    cout << ans[i] + i + 1 << "\n";
  }
}
