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>
long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
int n, a[150010], p[150010];
struct data {
  int mx, se, cnt;
} s[600000];
long long sum[600010], ans[150010];
int tag[600010], tag2[600010], cnt[600010];
data operator+(data a, const data& b) {
  if (b.mx == a.mx) return {a.mx, std::max(a.se, b.se), a.cnt + b.cnt};
  if (a.mx > b.mx) return {a.mx, std::max(a.se, b.mx), a.cnt};
  return {b.mx, std::max(b.se, a.mx), b.cnt};
}
void upd(int x, int y) {
  sum[x] += 1ll * cnt[x] * y, s[x].mx += y, s[x].se += y, tag[x] += y;
}
void upd2(int x, int y) {
  s[x].mx += y, sum[x] += 1ll * s[x].cnt * y, tag2[x] += y;
}
void down(int x) {
  if (tag[x]) upd(x << 1, tag[x]), upd(x << 1 | 1, tag[x]), tag[x] = 0;
  if (tag2[x]) {
    if (s[x << 1].mx + tag2[x] == s[x].mx) upd2(x << 1, tag2[x]);
    if (s[x << 1 | 1].mx + tag2[x] == s[x].mx) upd2(x << 1 | 1, tag2[x]);
    tag2[x] = 0;
  }
}
void update_set(int x, int l, int r, int p, int d) {
  if (l == r) {
    s[x] = {d, -1919810, 1};
    sum[x] = d;
    ++cnt[x];
    return;
  }
  down(x);
  if (p <= ((l + r) >> 1))
    update_set(x << 1, l, ((l + r) >> 1), p, d);
  else
    update_set(x << 1 | 1, ((l + r) >> 1) + 1, r, p, d);
  s[x] = s[x << 1] + s[x << 1 | 1];
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
  ++cnt[x];
}
void update_min(int x, int l, int r, int R, int y) {
  if (s[x].mx <= y) return;
  if (r <= R && s[x].se < y) return upd2(x, y - s[x].mx);
  down(x);
  update_min(x << 1, l, ((l + r) >> 1), R, y);
  if (((l + r) >> 1) < R) update_min(x << 1 | 1, ((l + r) >> 1) + 1, r, R, y);
  s[x] = s[x << 1] + s[x << 1 | 1];
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void update_add(int x, int l, int r, int L) {
  if (L <= l) return upd(x, 1);
  down(x);
  if (L <= ((l + r) >> 1)) update_add(x << 1, l, ((l + r) >> 1), L);
  update_add(x << 1 | 1, ((l + r) >> 1) + 1, r, L);
  s[x] = s[x << 1] + s[x << 1 | 1];
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
int query(int x, int l, int r, int p) {
  if (l == r) return sum[x];
  down(x);
  if (p <= ((l + r) >> 1))
    return query(x << 1, l, ((l + r) >> 1), p);
  else
    return query(x << 1 | 1, ((l + r) >> 1) + 1, r, p);
}
int tr[150010];
void update(int x) {
  while (x <= n) ++tr[x], x += x & -x;
}
int query(int x) {
  int r = 0;
  while (x) r += tr[x], x -= x & -x;
  return r;
}
void solve() {
  for (int i = 1; i <= n; ++i) p[a[i]] = i;
  memset(s, 0, sizeof s);
  memset(sum, 0, sizeof sum);
  memset(cnt, 0, sizeof cnt);
  memset(tr, 0, sizeof tr);
  memset(tag, 0, sizeof tag);
  memset(tag2, 0, sizeof tag2);
  for (int i = 1; i <= n; ++i) {
    update_set(1, 1, n, p[i], i);
    if (query(p[i])) update_min(1, 1, n, p[i] - 1, query(p[i]));
    if (p[i] < n) update_add(1, 1, n, p[i] + 1);
    ans[i] += sum[1];
    update(p[i]);
  }
}
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  solve();
  std::reverse(a + 1, a + n + 1);
  solve();
  for (int i = 1; i <= n; ++i) printf("%lld\n", ans[i] - 1ll * i * i);
  return 0;
}
