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 w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 1.5e5 + 10;
int n, p[maxn];
long long ans[maxn];
int tag[maxn << 2], ptag[maxn << 2], mx[maxn << 2], pmx[maxn << 2],
    s[maxn << 2], num[maxn << 2];
long long sum[maxn << 2];
void clear(int ro = 1, int l = 1, int r = n) {
  mx[ro] = pmx[ro] = num[ro] = tag[ro] = ptag[ro] = s[ro] = sum[ro] = 0;
  if (l == r) return;
  clear((ro << 1), l, ((l + r) >> 1)),
      clear((ro << 1 | 1), ((l + r) >> 1) + 1, r);
}
inline void pushup(int ro) {
  num[ro] = num[(ro << 1)] + num[(ro << 1 | 1)],
  sum[ro] = sum[(ro << 1)] + sum[(ro << 1 | 1)];
  if (mx[(ro << 1)] > mx[(ro << 1 | 1)])
    mx[ro] = mx[(ro << 1)], pmx[ro] = max(pmx[(ro << 1)], mx[(ro << 1 | 1)]),
    s[ro] = s[(ro << 1)];
  else if (mx[(ro << 1)] < mx[(ro << 1 | 1)])
    mx[ro] = mx[(ro << 1 | 1)],
    pmx[ro] = max(pmx[(ro << 1 | 1)], mx[(ro << 1)]), s[ro] = s[(ro << 1 | 1)];
  else
    mx[ro] = mx[(ro << 1)], pmx[ro] = max(pmx[(ro << 1)], pmx[(ro << 1 | 1)]),
    s[ro] = s[(ro << 1)] + s[(ro << 1 | 1)];
}
inline void addtag(int ro, int k, int pk) {
  mx[ro] += k, tag[ro] += k, pmx[ro] += pk, ptag[ro] += pk;
  sum[ro] += 1ll * k * s[ro] + 1ll * pk * (num[ro] - s[ro]);
}
inline void pushdown(int ro) {
  bool tp = mx[(ro << 1 | 1)] >= mx[(ro << 1)];
  addtag((ro << 1), mx[(ro << 1)] >= mx[(ro << 1 | 1)] ? tag[ro] : ptag[ro],
         ptag[ro]),
      addtag((ro << 1 | 1), tp ? tag[ro] : ptag[ro], ptag[ro]);
  tag[ro] = ptag[ro] = 0;
}
int add(int x, int y, int ro = 1, int l = 1, int r = n) {
  if (l > y or r < x) return 0;
  if (x <= l and y >= r) return addtag(ro, 1, 1), num[ro];
  pushdown(ro);
  int ans = add(x, y, (ro << 1), l, ((l + r) >> 1)) +
            add(x, y, (ro << 1 | 1), ((l + r) >> 1) + 1, r);
  pushup(ro);
  return ans;
}
void modify(int x, int k, int ro = 1, int l = 1, int r = n) {
  if (l == r) return sum[ro] = mx[ro] = k, s[ro] = num[ro] = 1, void();
  pushdown(ro);
  x <= ((l + r) >> 1) ? modify(x, k, (ro << 1), l, ((l + r) >> 1))
                      : modify(x, k, (ro << 1 | 1), ((l + r) >> 1) + 1, r);
  pushup(ro);
}
void updmn(int x, int y, int k, int ro = 1, int l = 1, int r = n) {
  if (k >= mx[ro] or l > y or r < x) return;
  if (x <= l and y >= r and pmx[ro] < k)
    return addtag(ro, min(k - mx[ro], 0), 0);
  pushdown(ro);
  updmn(x, y, k, (ro << 1), l, ((l + r) >> 1)),
      updmn(x, y, k, (ro << 1 | 1), ((l + r) >> 1) + 1, r);
  pushup(ro);
}
inline void work() {
  n = read();
  for (int i = 1; i <= n; i++) p[read()] = i;
  for (int x, i = 1; i <= n; i++)
    x = add(p[i] + 1, n), modify(p[i], i + 1), updmn(1, p[i] - 1, i - x),
    ans[i] += sum[1];
  for (int i = 1; i <= n; i++) p[i] = n - p[i] + 1;
  clear();
  for (int x, i = 1; i <= n; i++)
    x = add(p[i] + 1, n), modify(p[i], i + 1), updmn(1, p[i] - 1, i - x),
    ans[i] += sum[1];
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] -= 1ll * i * (i + 2));
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}
