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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 1.5e5 + 5;
int T[N], tg0[N << 2], tg1[N << 2], mn0[N << 2], mn1[N << 2], sz[N << 2],
    cnt[N << 2], n, p[N], a[N];
long long sum[N << 2], ans[N];
void up(int x) {
  while (x <= n) {
    T[x]++;
    x += (x & -x);
  }
}
int que(int x) {
  int res = 0;
  while (x) {
    res += T[x];
    x -= (x & -x);
  }
  return res;
}
void upd(int x, int v) {
  tg0[x] += v;
  tg1[x] += v;
  mn0[x] += v;
  mn1[x] += v;
  sum[x] += 1ll * sz[x] * v;
}
void mxd(int x, int v) {
  tg1[x] = max(tg1[x], v);
  if (mn0[x] >= v || !sz[x]) return;
  sum[x] += 1ll * cnt[x] * (v - mn0[x]);
  mn0[x] = v;
}
void pushup(int x) {
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
  sz[x] = sz[x << 1] + sz[x << 1 | 1];
  mn0[x] = min(mn0[x << 1], mn0[x << 1 | 1]);
  mn1[x] = min(mn0[x << 1] == mn0[x] ? mn1[x << 1] : mn0[x << 1],
               mn0[x << 1 | 1] == mn0[x] ? mn1[x << 1 | 1] : mn0[x << 1 | 1]);
  cnt[x] = (mn0[x] == mn0[x << 1] ? cnt[x << 1] : 0) +
           (mn0[x] == mn0[x << 1 | 1] ? cnt[x << 1 | 1] : 0);
}
void pushdown(int x) {
  if (tg0[x]) {
    upd(x << 1, tg0[x]);
    upd(x << 1 | 1, tg0[x]);
    tg0[x] = 0;
  }
  if (tg1[x]) {
    mxd(x << 1, tg1[x]);
    mxd(x << 1 | 1, tg1[x]);
    tg1[x] = 0;
  }
}
void upp(int x, int l, int r, int p) {
  if (l == r) return cnt[x] = sz[x] = 1, mn0[x] = 0, void();
  pushdown(x);
  int mid = l + r >> 1;
  if (p <= mid)
    upp(x << 1, l, mid, p);
  else
    upp(x << 1 | 1, mid + 1, r, p);
  pushup(x);
}
void up1(int x, int l, int r, int p, int q) {
  if (p <= l && r <= q) return upd(x, 1);
  pushdown(x);
  int mid = l + r >> 1;
  if (p <= mid) up1(x << 1, l, mid, p, q);
  if (q > mid) up1(x << 1 | 1, mid + 1, r, p, q);
  pushup(x);
}
void upx(int x, int l, int r, int p, int q, int v) {
  if (mn0[x] >= v || !sz[x]) return;
  if (p <= l && r <= q && mn1[x] > v) return mxd(x, v);
  pushdown(x);
  int mid = l + r >> 1;
  if (p <= mid) upx(x << 1, l, mid, p, q, v);
  if (q > mid) upx(x << 1 | 1, mid + 1, r, p, q, v);
  pushup(x);
}
void doit() {
  memset(sum, 0, sizeof sum);
  memset(tg0, 0, sizeof tg0);
  memset(tg1, 0, sizeof tg1);
  memset(mn0, 0x3f, sizeof mn0);
  memset(mn1, 0x3f, sizeof mn1);
  memset(cnt, 0, sizeof cnt);
  memset(sz, 0, sizeof sz);
  for (int i = 1; i <= n; i++) T[i] = 0;
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    int p = ::p[i];
    up(p);
    int v = que(p);
    upp(1, 1, n, p);
    if (p < n) {
      up1(1, 1, n, p + 1, n);
      upx(1, 1, n, p + 1, n, v);
    }
    ans[i] += 1ll * i * (i - 1) / 2 - sum[1];
  }
}
signed main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  doit();
  reverse(a + 1, a + 1 + n);
  doit();
  for (int i = 1; i <= n; i++) write(i + ans[i]), puts("");
}
