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;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x;
}
const int INF = 0x3f3f3f3f;
const int MAXN = 150000 + 233;
int n;
struct node {
  int mx1, mx2, cnt;
  node() { mx1 = mx2 = cnt = 0; }
  node(int A, int B, int C) { mx1 = A, mx2 = B, cnt = C; }
  node operator+(node B) {
    node A = *this;
    if (A.mx1 == B.mx1) {
      return node(A.mx1, max(A.mx2, B.mx2), A.cnt + B.cnt);
    }
    if (A.mx1 < B.mx1) {
      return node(B.mx1, max(A.mx1, B.mx2), B.cnt);
    } else {
      return node(A.mx1, max(A.mx2, B.mx1), A.cnt);
    }
  }
};
const int N = 1 << 18;
long long rest = 0;
struct SGT_BEATS {
  node tree[N + N];
  int siz[N + N];
  SGT_BEATS() { memset(siz, 0, sizeof(siz)); }
  void add(int pos) {
    while (pos <= n) {
      siz[pos]++;
      pos += pos & (-pos);
    }
  }
  int sum(int pos) {
    int s = 0;
    while (pos) {
      s += siz[pos];
      pos -= pos & (-pos);
    }
    return s;
  }
  void pu(int now) { tree[now] = tree[now << 1] + tree[now << 1 | 1]; }
  void pd(int now) {
    if (now < N) {
      tree[now << 1].mx1 = min(tree[now << 1].mx1, tree[now].mx1);
      tree[now << 1 | 1].mx1 = min(tree[now << 1 | 1].mx1, tree[now].mx1);
    }
  }
  void modify(int pos, int now = 1, int l = 1, int r = N + 1) {
    pd(now);
    if (l == r - 1) {
      tree[now] = node{n + 1, 0, 1};
      return;
    }
    int mid = (l + r) >> 1;
    if (mid > pos)
      modify(pos, now << 1, l, mid);
    else
      modify(pos, now << 1 | 1, mid, r);
    pu(now);
  }
  void upd(int pos, int now = 1, int l = 1, int r = N + 1) {
    pd(now);
    if (l >= pos) return;
    if (r <= pos) {
      if (tree[now].mx1 <= pos) {
        return;
      }
      if (tree[now].mx2 < pos) {
        rest -= 1ll * tree[now].cnt * (sum(tree[now].mx1 - 1) - sum(pos - 1));
        tree[now].mx1 = pos;
        pd(now);
        return;
      }
    }
    int mid = (l + r) >> 1;
    upd(pos, now << 1, l, mid);
    upd(pos, now << 1 | 1, mid, r);
    pu(now);
  }
} sgt1, sgt2;
int a[MAXN];
int pos[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) a[i] = read(), pos[a[i]] = i;
  for (int i = 1; i <= n; ++i) {
    sgt1.modify(pos[i]);
    sgt2.modify(n - pos[i] + 1);
    sgt1.upd(pos[i]);
    sgt2.upd(n - pos[i] + 1);
    sgt1.add(pos[i]);
    sgt2.add(n - pos[i] + 1);
    rest += i;
    printf("%lld\n", rest);
  }
  return 0;
}
