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 INF = 2147483647;
struct Tree {
  int mx, t, se;
  long long num, sum;
  int add, val;
} tree[600005];
int n, m, a[300005], pos[300005];
long long ans[150005];
void pushup(int x) {
  tree[x].mx = max(tree[x << 1].mx, tree[x << 1 | 1].mx);
  tree[x].se = tree[x << 1].mx == tree[x << 1 | 1].mx
                   ? max(tree[x << 1].se, tree[x << 1 | 1].se)
                   : max(min(tree[x << 1].mx, tree[x << 1 | 1].mx),
                         max(tree[x << 1].se, tree[x << 1 | 1].se));
  tree[x].t = 0;
  if (tree[x << 1].mx == tree[x].mx) tree[x].t += tree[x << 1].t;
  if (tree[x << 1 | 1].mx == tree[x].mx) tree[x].t += tree[x << 1 | 1].t;
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
  tree[x].num = tree[x << 1].num + tree[x << 1 | 1].num;
}
void pushdown(int x) {
  if (tree[x].add) {
    int v = tree[x].add;
    if (tree[x << 1].num) {
      tree[x << 1].add += v, tree[x << 1].mx += v;
      if (tree[x << 1].val != INF) tree[x << 1].val += v;
      if (tree[x << 1].se != -INF) tree[x << 1].se += v;
      tree[x << 1].sum += v * tree[x << 1].num;
    }
    if (tree[x << 1 | 1].num) {
      tree[x << 1 | 1].add += v, tree[x << 1 | 1].mx += v;
      if (tree[x << 1 | 1].val != INF) tree[x << 1 | 1].val += v;
      if (tree[x << 1 | 1].se != -INF) tree[x << 1 | 1].se += v;
      tree[x << 1 | 1].sum += v * tree[x << 1 | 1].num;
    }
    tree[x].add = 0;
  }
  if (tree[x].val != INF) {
    int v = tree[x].val;
    if (tree[x << 1].num) {
      tree[x << 1].val = min(tree[x << 1].val, v);
      if (tree[x << 1].val < tree[x << 1].mx) {
        tree[x << 1].sum -=
            (tree[x << 1].mx - tree[x << 1].val) * tree[x << 1].t;
        tree[x << 1].mx = tree[x << 1].val;
      }
    }
    if (tree[x << 1 | 1].num) {
      tree[x << 1 | 1].val = min(tree[x << 1 | 1].val, v);
      if (tree[x << 1 | 1].val < tree[x << 1 | 1].mx) {
        tree[x << 1 | 1].sum -=
            (tree[x << 1 | 1].mx - tree[x << 1 | 1].val) * tree[x << 1 | 1].t;
        tree[x << 1 | 1].mx = tree[x << 1 | 1].val;
      }
    }
    tree[x].val = INF;
  }
}
void build(int x, int l, int r) {
  if (l == r) {
    tree[x].num = tree[x].sum = tree[x].add = tree[x].mx = tree[x].t = 0;
    tree[x].se = -INF;
    tree[x].val = INF;
    return;
  }
  build(x << 1, l, ((l + r) >> 1));
  build(x << 1 | 1, ((l + r) >> 1) + 1, r);
  pushup(x);
}
void update(int x, int l, int r, int p) {
  if (l == r) {
    tree[x].mx = m;
    tree[x].num = 1;
    tree[x].sum = m;
    tree[x].t = 1;
    return;
  }
  pushdown(x);
  if (p <= ((l + r) >> 1))
    update(x << 1, l, ((l + r) >> 1), p);
  else
    update(x << 1 | 1, ((l + r) >> 1) + 1, r, p);
  pushup(x);
}
void add(int x, int l, int r, int from, int to, int v) {
  if (from > to) return;
  if (!tree[x].num) return;
  if (l >= from && r <= to) {
    tree[x].add += v;
    if (tree[x].val != INF) tree[x].val += v;
    tree[x].mx += v;
    if (tree[x].se != -INF) tree[x].se += v;
    tree[x].sum += tree[x].num * v;
    return;
  }
  pushdown(x);
  if (from <= ((l + r) >> 1)) add(x << 1, l, ((l + r) >> 1), from, to, v);
  if (to > ((l + r) >> 1)) add(x << 1 | 1, ((l + r) >> 1) + 1, r, from, to, v);
  pushup(x);
}
void setmax(int x, int l, int r, int from, int to, int v) {
  if (from > to) return;
  if (!tree[x].num) return;
  if (l >= from && r <= to) {
    if (v >= tree[x].mx) return;
    if (v > tree[x].se) {
      tree[x].sum -= (tree[x].mx - v) * tree[x].t;
      tree[x].mx = v;
      tree[x].val = v;
      return;
    }
    pushdown(x);
    setmax(x << 1, l, ((l + r) >> 1), from, to, v);
    setmax(x << 1 | 1, ((l + r) >> 1) + 1, r, from, to, v);
    pushup(x);
    return;
  }
  pushdown(x);
  if (from <= ((l + r) >> 1)) setmax(x << 1, l, ((l + r) >> 1), from, to, v);
  if (to > ((l + r) >> 1))
    setmax(x << 1 | 1, ((l + r) >> 1) + 1, r, from, to, v);
  pushup(x);
}
int c[150005];
void Add(int p) {
  while (p <= n) {
    c[p]++;
    p += p ^ (p & (p - 1));
  }
}
int Sum(int p) {
  int ans = 0;
  while (p) {
    ans += c[p];
    p &= p - 1;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  build(1, 1, n);
  for (m = 1; m <= n; m++) {
    setmax(1, 1, n, 1, pos[m] - 1, Sum(pos[m]));
    Add(pos[m]);
    add(1, 1, n, pos[m], n, 1);
    update(1, 1, n, pos[m]);
    ans[m] += tree[1].sum;
  }
  memset(c, 0, sizeof(c));
  for (int i = 1; i * 2 <= n; i++) swap(a[i], a[n - i + 1]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  build(1, 1, n);
  for (m = 1; m <= n; m++) {
    setmax(1, 1, n, 1, pos[m] - 1, Sum(pos[m]));
    Add(pos[m]);
    add(1, 1, n, pos[m], n, 1);
    update(1, 1, n, pos[m]);
    ans[m] -= tree[1].num * m - tree[1].sum;
  }
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
  return 0;
}
