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, i, x, a[N], ia[N];
long long ans;
struct atom {
  int m0, m1, c;
};
inline atom operator+(atom A, atom B) {
  if (A.m0 < B.m0) swap(A, B);
  return A.m0 == B.m0 ? (atom){A.m0, max(A.m1, B.m1), A.c + B.c}
                      : (atom){A.m0, max(A.m1, B.m0), A.c};
}
struct T {
  int bi[N];
  inline void add(int x) {
    for (; x <= n; x += x & -x) ++bi[x];
  }
  inline int ask(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += bi[x];
    return ans;
  }
  static const int M = N << 2;
  atom t[M];
  int ov[M];
  inline void fix(int x, int y) {
    if (t[x].m0 > t[y].m0) t[x].m0 = t[y].m0;
  }
  inline void pushdown(int i) {
    if (ov[i] > t[i].m0) fix(i << 1, i), fix(i << 1 | 1, i), ov[i] = t[i].m0;
  }
  inline void upd(int i) {
    t[i] = t[i << 1] + t[i << 1 | 1];
    ov[i] = t[i].m0;
  }
  void mdy(int i, int l, int r, int ql, int qr, int v) {
    int m = l + r >> 1;
    if (ql <= l && r <= qr) {
      if (v >= t[i].m0) return;
      if (v < t[i].m1)
        pushdown(i), mdy(i << 1, l, m, ql, qr, v),
            mdy(i << 1 | 1, m + 1, r, ql, qr, v), upd(i);
      else
        ans -= 1ll * t[i].c * (ask(t[i].m0) - ask(v)), t[i].m0 = v;
      return;
    }
    pushdown(i);
    if (ql <= m) mdy(i << 1, l, m, ql, qr, v);
    if (qr > m) mdy(i << 1 | 1, m + 1, r, ql, qr, v);
    upd(i);
  }
  void set(int i, int k, int l, int r) {
    if (l == k && r == k) {
      t[i] = (atom){n, 0, 1};
      return;
    }
    int m = l + r >> 1;
    pushdown(i);
    if (k <= m)
      set(i << 1, k, l, m);
    else
      set(i << 1 | 1, k, m + 1, r);
    upd(i);
  }
} A, B;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; ++i) cin >> a[i], ia[a[i]] = i;
  for (i = 1; i < T::M; ++i) A.ov[i] = B.ov[i] = N;
  for (i = 1; i <= n; ++i) {
    x = ia[i];
    if (x > 1) A.mdy(1, 1, n, 1, x - 1, x - 1);
    if (x < n) B.mdy(1, 1, n, 1, n - x, n - x);
    A.add(x);
    B.add(n - x + 1);
    A.set(1, x, 1, n);
    B.set(1, n - x + 1, 1, n);
    ans += i;
    cout << ans << '\n';
  }
  return 0;
}
