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>
const int N = 2e5;
const int inf = 1e9;
using namespace std;
int n, a[N + 5], pos[N + 5], s[N + 5];
long long ans[N + 5];
struct node {
  int mx, mxi, smx, s;
  long long sm;
  node operator+(const node &a) const {
    node c = {0, 0, 0, 0, 0};
    c.sm = sm + a.sm;
    c.s = s + a.s;
    if (mx > a.mx) {
      c.mx = mx;
      c.smx = smx;
      c.mxi = max(a.mx, mxi);
    }
    if (mx < a.mx) {
      c.mx = a.mx;
      c.smx = a.smx;
      c.mxi = max(a.mxi, mx);
    }
    if (mx == a.mx) {
      c.mx = mx;
      c.smx = a.smx + smx;
      c.mxi = max(a.mxi, mxi);
    }
    return c;
  }
};
struct Seg {
  node s[N * 4 + 5];
  int tag[N * 4 + 5], ct[N * 4 + 5];
  void build(int k, int l, int r) {
    s[k] = {0, -inf, 0, 0, 0};
    tag[k] = 0;
    ct[k] = inf;
    if (l == r) return;
    int mid = l + r >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
  }
  void add(int k, int l, int r, int z) {
    if (!s[k].s) return;
    s[k].sm += 1ll * s[k].s * z;
    s[k].mx += z;
    if (s[k].mxi != -inf) s[k].mxi += z;
    tag[k] += z;
    if (ct[k] != inf) ct[k] += z;
  }
  void cha(int k, int z) {
    if (!s[k].s) return;
    if (z >= s[k].mx) return;
    s[k].sm -= 1ll * s[k].smx * (s[k].mx - z);
    s[k].mx = z;
    ct[k] = z;
  }
  void pushdown(int k, int l, int r, int mid) {
    if (tag[k]) {
      add(k << 1, l, mid, tag[k]);
      add(k << 1 | 1, mid + 1, r, tag[k]);
      tag[k] = 0;
    }
    if (ct[k] != inf) {
      cha(k << 1, ct[k]);
      cha(k << 1 | 1, ct[k]);
      ct[k] = inf;
    }
  }
  void insert(int k, int l, int r, int x, int z) {
    if (l == r) {
      s[k] = (node){z, -inf, 1, 1, z};
      return;
    }
    int mid = l + r >> 1;
    pushdown(k, l, r, mid);
    if (x <= mid)
      insert(k << 1, l, mid, x, z);
    else
      insert(k << 1 | 1, mid + 1, r, x, z);
    s[k] = s[k << 1] + s[k << 1 | 1];
  }
  void modify(int k, int l, int r, int x, int y, int z) {
    if (l >= x && r <= y) {
      add(k, l, r, z);
      return;
    }
    int mid = l + r >> 1;
    pushdown(k, l, r, mid);
    if (x <= mid) modify(k << 1, l, mid, x, y, z);
    if (y > mid) modify(k << 1 | 1, mid + 1, r, x, y, z);
    s[k] = s[k << 1] + s[k << 1 | 1];
  }
  void getmin(int k, int l, int r, int x, int y, int z) {
    if (l >= x && r <= y) {
      if (z > s[k].mxi) {
        cha(k, z);
        return;
      }
    }
    int mid = l + r >> 1;
    pushdown(k, l, r, mid);
    if (x <= mid) getmin(k << 1, l, mid, x, y, z);
    if (y > mid) getmin(k << 1 | 1, mid + 1, r, x, y, z);
    s[k] = s[k << 1] + s[k << 1 | 1];
  }
  int query(int k, int l, int r, int x, int y) {
    if (l >= x && r <= y) return s[k].s;
    int mid = l + r >> 1;
    pushdown(k, l, r, mid);
    if (y <= mid) return query(k << 1, l, mid, x, y);
    if (x > mid) return query(k << 1 | 1, mid + 1, r, x, y);
    return query(k << 1, l, mid, x, y) + query(k << 1 | 1, mid + 1, r, x, y);
  }
} tree;
void solve() {
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  tree.build(1, 1, n);
  memset(s, 0, sizeof(s));
  for (int k = 1; k <= n; k++) {
    if (pos[k] + 1 <= n) tree.modify(1, 1, n, pos[k] + 1, n, 1);
    if (pos[k] - 1 >= 1) {
      int x = tree.query(1, 1, n, 1, pos[k] - 1) + 1;
      tree.getmin(1, 1, n, 1, pos[k] - 1, x);
    }
    tree.insert(1, 1, n, pos[k], k + 1);
    ans[k] += tree.s[1].sm;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve();
  reverse(a + 1, a + n + 1);
  solve();
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] - 1ll * i * (i + 2));
  return 0;
}
