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 MAXN = 1.5e5 + 5;
const int inf = 0x3f3f3f3f;
inline void chk_max(int &a, int b) {
  if (a < b) a = b;
}
struct BIT {
  int tree[MAXN], n;
  inline void clear(int _n) {
    n = _n;
    memset(tree, 0, (n + 1) << 2);
  }
  inline void update(int x, int k) {
    while (x <= n) tree[x] += k, x += ((x) & (-x));
  }
  inline int query(int x) {
    int res = 0;
    while (x) res += tree[x], x ^= ((x) & (-x));
    return res;
  }
  inline int query(int l, int r) { return query(r) - query(l - 1); }
};
struct Segment_Tree {
  int fir[MAXN << 2], sec[MAXN << 2], fcnt[MAXN << 2], tag[MAXN << 2];
  long long sum[MAXN << 2];
  inline void set_mx(int u, int _fir, int _sec, int _fcnt) {
    fir[u] = _fir;
    sec[u] = _sec;
    fcnt[u] = _fcnt;
  }
  inline void push_up(int u) {
    int l = ((u) << 1), r = ((u) << 1 | 1);
    sum[u] = sum[l] + sum[r];
    if (fir[l] < fir[r]) swap(l, r);
    if (fir[l] == fir[r])
      set_mx(u, fir[l], max(sec[l], sec[r]), fcnt[l] + fcnt[r]);
    else
      set_mx(u, fir[l], max(sec[l], fir[r]), fcnt[l]);
  }
  inline void upd_add(int u, int len, int k) {
    tag[u] += k;
    fir[u] += k;
    sec[u] += k;
    sum[u] += (long long)len * k;
  }
  inline void push_down(int u, int l, int r) {
    if (tag[u]) {
      int mid = (l + r) >> 1;
      upd_add(((u) << 1), mid - l + 1, tag[u]);
      upd_add(((u) << 1 | 1), r - mid, tag[u]);
      tag[u] = 0;
    }
    if (fir[((u) << 1)] > fir[u])
      sum[((u) << 1)] +=
          (long long)(fir[u] - fir[((u) << 1)]) * fcnt[((u) << 1)],
          fir[((u) << 1)] = fir[u];
    if (fir[((u) << 1 | 1)] > fir[u])
      sum[((u) << 1 | 1)] +=
          (long long)(fir[u] - fir[((u) << 1 | 1)]) * fcnt[((u) << 1 | 1)],
          fir[((u) << 1 | 1)] = fir[u];
  }
  void build(int u, int l, int r) {
    tag[u] = 0;
    if (l == r) {
      fir[u] = 0;
      sec[u] = -inf;
      fcnt[u] = 1;
      sum[u] = 0;
      return;
    }
    int mid = (l + r) >> 1;
    build(((u) << 1), l, mid);
    build(((u) << 1 | 1), mid + 1, r);
    push_up(u);
  }
  void update_add(int u, int l, int r, int ql, int qr, int k) {
    if (ql <= l && r <= qr) {
      upd_add(u, r - l + 1, k);
      return;
    }
    push_down(u, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) update_add(((u) << 1), l, mid, ql, qr, k);
    if (mid < qr) update_add(((u) << 1 | 1), mid + 1, r, ql, qr, k);
    push_up(u);
  }
  void update_min(int u, int l, int r, int ql, int qr, int k) {
    if (fir[u] <= k) return;
    if (ql <= l && r <= qr && sec[u] < k) {
      sum[u] += (long long)(k - fir[u]) * fcnt[u];
      fir[u] = k;
      return;
    }
    push_down(u, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) update_min(((u) << 1), l, mid, ql, qr, k);
    if (mid < qr) update_min(((u) << 1 | 1), mid + 1, r, ql, qr, k);
    push_up(u);
  }
};
int n;
int a[MAXN];
vector<long long> get_res(void) {
  static int pos[MAXN];
  for (int i = 1; i <= n; ++i) pos[a[i]] = i;
  vector<long long> res(n + 1);
  static BIT num;
  static Segment_Tree tree;
  num.clear(n);
  tree.build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    num.update(pos[i], 1);
    int k = num.query(pos[i]);
    tree.update_add(1, 1, n, pos[i], pos[i], i - k + 1);
    if (pos[i] > 1) tree.update_min(1, 1, n, 1, pos[i] - 1, k - 1);
    if (pos[i] < n) tree.update_add(1, 1, n, pos[i] + 1, n, 1);
    res[i] = tree.sum[1];
  }
  return res;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  vector<long long> r = get_res();
  reverse(a + 1, a + n + 1);
  vector<long long> l = get_res();
  for (int i = 1; i <= n; ++i)
    printf("%lld\n", r[i] - ((long long)n * i - l[i]));
  return 0;
}
