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 long long N = 2e6 + 18;
long long n, i, t, a[N];
struct xdsl {
  long long mn[N], mn1[N], cnt[N], cv[N], sum[N], si[N], ad[N];
  void cov1(long long k, long long u) {
    if (si[k])
      if (mn[k] < u) {
        sum[k] += cnt[k] * (u - mn[k]), mn[k] = u, cv[k] = u;
      }
  }
  void add(long long k, long long u) {
    if (!si[k]) return;
    if (cv[k]) cv[k] += u;
    sum[k] += si[k] * u;
    ad[k] += u, mn[k] += u, mn1[k] += u;
  }
  void cov(long long k, long long u) {
    if (!si[k]) return;
    if (ad[k]) add(k << 1, ad[k]), add(k << 1 | 1, ad[k]), ad[k] = 0;
    if (mn[k] < u) {
      sum[k] += cnt[k] * (u - mn[k]), mn[k] = u, cv[k] = u;
    }
  }
  void PD(long long k) {
    if (ad[k]) add(k << 1, ad[k]), add(k << 1 | 1, ad[k]), ad[k] = 0;
    if (cv[k]) cov(k << 1, cv[k]), cov(k << 1 | 1, cv[k]), cv[k] = 0;
  }
  void PU(long long k) {
    long long lc = k << 1, rc = lc + 1;
    sum[k] = sum[lc] + sum[rc];
    si[k] = si[lc] + si[rc];
    if (mn[lc] == mn[rc])
      mn[k] = mn[lc], cnt[k] = cnt[lc] + cnt[rc],
      mn1[k] = min(mn1[lc], mn1[rc]);
    else if (mn[lc] < mn[rc])
      mn[k] = mn[lc], cnt[k] = cnt[lc], mn1[k] = min(mn1[lc], mn[rc]);
    else
      mn[k] = mn[rc], cnt[k] = cnt[rc], mn1[k] = min(mn[lc], mn1[rc]);
  }
  void C(long long k, long long l, long long r, long long u, long long v) {
    PD(k);
    if (si[k])
      if (u <= l) {
        if (mn1[k] <= v)
          C(k << 1, l, (l + r) >> 1, u, v),
              C(k << 1 | 1, ((l + r) >> 1) + 1, r, u, v), PU(k);
        else if (mn[k] < v)
          cov(k, v);
      } else {
        if (u <= (l + r) / 2) C(k << 1, l, (l + r) >> 1, u, v);
        C(k << 1 | 1, ((l + r) >> 1) + 1, r, u, v), PU(k);
      }
  }
  void A(long long k, long long l, long long r, long long u) {
    PD(k);
    if (l == r) {
      cnt[k] = si[k] = 1;
      mn[k] = 0;
      return;
    }
    if (u <= (l + r) / 2)
      A(k << 1, l, (l + r) >> 1, u);
    else
      A(k << 1 | 1, ((l + r) >> 1) + 1, r, u);
    PU(k);
  }
  void AA(long long k, long long l, long long r, long long u) {
    if (!si[k]) return;
    PD(k);
    if (u <= l) {
      add(k, 1);
      return;
    }
    if (u <= (l + r) / 2) AA(k << 1, l, (l + r) >> 1, u);
    AA(k << 1 | 1, ((l + r) >> 1) + 1, r, u), PU(k);
  }
  long long Q(long long k, long long l, long long r, long long u) {
    PD(k);
    if (r <= u) return si[k];
    long long e = Q(k << 1, l, (l + r) >> 1, u);
    if ((l + r) / 2 < u) e += Q(k << 1 | 1, ((l + r) >> 1) + 1, r, u);
    return e;
  }
} l, r;
signed main() {
  scanf("%lld", &n);
  for (i = (1); i <= (n); i++) scanf("%lld", &t), a[t] = i;
  for (i = (1); i <= (n * 4); i++)
    l.mn1[i] = r.mn1[i] = l.mn[i] = r.mn[i] = 20000000;
  for (i = (1); i <= (n); i++) {
    t = l.Q(1, 1, n, a[i]) + 1;
    if (i == 7) l.PD(6), l.PU(6);
    l.AA(1, 1, n, a[i]);
    l.C(1, 1, n, a[i], t);
    l.A(1, 1, n, a[i]);
    r.AA(1, 1, n, n - a[i] + 1);
    r.C(1, 1, n, n - a[i] + 1, i - t + 1);
    r.A(1, 1, n, n - a[i] + 1);
    printf("%lld\n", (i + 1) * i - r.sum[1] - l.sum[1] - i);
  }
  return 0;
}
