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 = 150000 + 9;
int a[N], p[N], k;
long long tot, ans[N];
template <int N, typename T>
struct Binary_Indexed_Tree {
  int n;
  T s[N];
  void init(int _n) {
    n = _n;
    memset(s, 0, sizeof(*s));
  }
  void upd_add(int p, T k) {
    for (; p <= n; p += p & -p) s[p] += k;
  }
  T q_sum(int p) {
    T r = 0;
    for (; p; p -= p & -p) r += s[p];
    return r;
  }
};
Binary_Indexed_Tree<N, int> bit[2];
template <int N, typename T>
struct Segment_Tree {
  int n, L, R, cnt1[N * 4];
  T *A, K, max1[N * 4], max2[N * 4], sum[N * 4];
  void push_down(int o, int lc, int rc) {
    if (max1[o] < max1[lc]) {
      max1[lc] = max1[o];
    }
    if (max1[o] < max1[rc]) {
      max1[rc] = max1[o];
    }
  }
  void push_up(int o, int lc, int rc) {
    if (max1[lc] > max1[rc]) {
      max1[o] = max1[lc];
      cnt1[o] = cnt1[lc];
      max2[o] = max(max2[lc], max1[rc]);
    } else if (max1[rc] > max1[lc]) {
      max1[o] = max1[rc];
      cnt1[o] = cnt1[rc];
      max2[o] = max(max2[rc], max1[lc]);
    } else {
      max1[o] = max1[lc];
      cnt1[o] = cnt1[lc] + cnt1[rc];
      max2[o] = max(max2[lc], max2[rc]);
    }
  }
  void __build(int o, int l, int r) {
    if (l == r) {
      max1[o] = 0;
      cnt1[o] = 0;
      max2[o] = INT_MIN;
    } else {
      int lc = o << 1, rc = lc | 1, m = (l + r) >> 1;
      __build(lc, l, m);
      __build(rc, m + 1, r);
      push_up(o, lc, rc);
    }
  }
  void build(T _A[], int _n) {
    A = _A;
    n = _n;
    __build(1, 1, n);
  }
  void __upd_min(int o, int l, int r) {
    if (l == r || (L <= l && r <= R && K > max2[o])) {
      if (K < max1[o]) {
        tot -= (long long)(bit[k].q_sum(max1[o]) - bit[k].q_sum(K)) * cnt1[o];
        max1[o] = K;
      }
    } else {
      int lc = o << 1, rc = lc | 1, m = (l + r) >> 1;
      push_down(o, lc, rc);
      if (L <= m) __upd_min(lc, l, m);
      if (R > m) __upd_min(rc, m + 1, r);
      push_up(o, lc, rc);
    }
  }
  void upd_min(int _L, int _R, int _K) {
    L = _L;
    R = _R;
    K = _K;
    __upd_min(1, 1, n);
  }
  void __upd_cnt(int o, int l, int r) {
    if (l == r) {
      max1[o] = n;
      cnt1[o] = 1;
    } else {
      int lc = o << 1, rc = lc | 1, m = (l + r) >> 1;
      push_down(o, lc, rc);
      if (L <= m)
        __upd_cnt(lc, l, m);
      else
        __upd_cnt(rc, m + 1, r);
      push_up(o, lc, rc);
    }
  }
  void upd_cnt(int _P) {
    L = _P;
    __upd_cnt(1, 1, n);
  }
};
Segment_Tree<N, int> segt[2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  bit[0].init(n);
  bit[1].init(n);
  segt[0].build(a, n);
  segt[1].build(a, n);
  for (int i = 1; i <= n; ++i) {
    k = 0;
    segt[k].upd_min(1, n - p[i] + 1, n - p[i]);
    segt[k].upd_cnt(n - p[i] + 1);
    bit[k].upd_add(n - p[i] + 1, 1);
    k = 1;
    segt[k].upd_min(1, p[i], p[i] - 1);
    segt[k].upd_cnt(p[i]);
    bit[k].upd_add(p[i], 1);
    tot += i;
    ans[i] = tot;
  }
  for (int i = 1; i <= n; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}
