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 = 15e4 + 1, INF = 0x3f3f3f3f;
long long int ans[N];
int n, a[N], b[N], c[N];
inline int lowbit(int x) { return x & -x; }
inline void add(int x) {
  while (x <= n) c[x]++, x += lowbit(x);
}
inline int ask(int x) { return !x ? 0 : (c[x] + ask(x ^ lowbit(x))); }
struct ele {
  struct node {
    int val, cnt;
    inline void set(int x, int y) { val = x, cnt = y; }
    inline bool operator<(const node& u) const { return val < u.val; }
    node operator+(const node& u) const {
      return (val == u.val) ? (node){val, cnt + u.cnt} : max(*this, u);
    }
  } x, y;
  long long int sum;
  int tag_max, tag_sum, siz;
  inline int val() { return x.val; }
  inline void add_max(int w) {
    sum += 1ll * w * x.cnt, x.val += w, tag_max += w;
  }
  inline void add_sum(int w) {
    sum += 1ll * w * siz, tag_sum += w;
    if (x.cnt) x.val += w;
    if (y.cnt) y.val += w;
  }
  inline void clean() {
    sum = tag_max = tag_sum = siz = 0, x.set(0, 0), y.set(-INF, 0);
  }
} t[N << 2];
inline void upd(int p) {
  t[p].sum = t[p << 1].sum + t[p << 1 | 1].sum,
  t[p].x = t[p << 1].x + t[p << 1 | 1].x;
  if (t[p << 1].val() == t[p << 1 | 1].val())
    t[p].y = t[p << 1].y + t[p << 1 | 1].y;
  else
    t[p].y = t[p << 1].y + t[p << 1 | 1].y + min(t[p << 1].x, t[p << 1 | 1].x);
}
inline void psd(int p) {
  int tmp = max(t[p << 1].val(), t[p << 1 | 1].val());
  if (t[p << 1].val() == tmp) t[p << 1].add_max(t[p].tag_max);
  if (t[p << 1 | 1].val() == tmp) t[p << 1 | 1].add_max(t[p].tag_max);
  t[p << 1].add_sum(t[p].tag_sum), t[p << 1 | 1].add_sum(t[p].tag_sum);
  t[p].tag_max = t[p].tag_sum = 0;
}
inline void input() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
}
inline void build(int p, int L, int R) {
  t[p].clean();
  if (L == R) return;
  int mid = L + R >> 1;
  build(p << 1, L, mid), build(p << 1 | 1, mid + 1, R), upd(p);
}
inline void dfs(int p, int x) {
  if (t[p].val() <= x) return;
  if (t[p].y.val < x) return t[p].add_max(x - t[p].val());
  psd(p), dfs(p << 1, x), dfs(p << 1 | 1, x), upd(p);
}
inline void modify(int p, int L, int R, int l, int r, int x) {
  if (l == L && R == r) return dfs(p, x);
  int mid = L + R >> 1;
  psd(p);
  if (r <= mid)
    modify(p << 1, L, mid, l, r, x);
  else if (l > mid)
    modify(p << 1 | 1, mid + 1, R, l, r, x);
  else
    modify(p << 1, L, mid, l, mid, x),
        modify(p << 1 | 1, mid + 1, R, mid + 1, r, x);
  upd(p);
}
inline void ins(int p, int L, int R, int l, int r, int x, int flag) {
  t[p].siz += flag;
  if (l == L && R == r) {
    if (flag) t[p].x.cnt = 1;
    return t[p].add_sum(x);
  }
  int mid = L + R >> 1;
  psd(p);
  if (r <= mid)
    ins(p << 1, L, mid, l, r, x, flag);
  else if (l > mid)
    ins(p << 1 | 1, mid + 1, R, l, r, x, flag);
  else
    ins(p << 1, L, mid, l, mid, x, flag),
        ins(p << 1 | 1, mid + 1, R, mid + 1, r, x, flag);
  upd(p);
}
inline void init() {
  build(1, 1, n), memset(c, 0, n + 1 << 2);
  for (int i = 1; i <= n; i++) b[a[i]] = i;
}
inline void solve() {
  init();
  for (int i = 1; i <= n; i++) {
    modify(1, 1, n, 1, b[i], ask(b[i])), ins(1, 1, n, b[i], b[i], i - 1, 1);
    ins(1, 1, n, b[i], n, 1, 0), add(b[i]), ans[i] += t[1].sum;
  }
}
inline void work() {
  solve(), reverse(a + 1, a + 1 + n), solve();
  for (int i = 1; i <= n; i++) cout << ans[i] - 1ll * i * i << '\n';
}
int main() {
  input();
  work();
  return 0;
}
