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 = 150005;
const int MAXM = (MAXN << 2);
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
int n, a[MAXN], pos[MAXN];
int mx[MAXM], mx2[MAXM], cnt[MAXM], siz[MAXM];
long long sum[MAXM], tag1[MAXM], tag2[MAXM];
long long ans[MAXN];
inline void puttag1(int rt, long long tg) {
  sum[rt] += tg * cnt[rt];
  if (siz[rt]) mx[rt] += tg;
  tag1[rt] += tg;
}
inline void puttag2(int rt, long long tg) {
  sum[rt] += tg * siz[rt];
  if (siz[rt]) mx[rt] += tg;
  if (siz[rt] > cnt[rt]) mx2[rt] += tg;
  tag2[rt] += tg;
}
inline void pushdown(int rt) {
  if (tag1[rt]) {
    int tmp = max(mx[(rt << 1)], mx[(rt << 1) + 1]);
    if (mx[(rt << 1)] == tmp) {
      puttag1((rt << 1), tag1[rt]);
    }
    if (mx[(rt << 1) + 1] == tmp) {
      puttag1((rt << 1) + 1, tag1[rt]);
    }
    tag1[rt] = 0;
  }
  if (tag2[rt]) {
    puttag2((rt << 1), tag2[rt]);
    puttag2((rt << 1) + 1, tag2[rt]);
    tag2[rt] = 0;
  }
}
inline void pushup(int rt) {
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1) + 1];
  siz[rt] = siz[(rt << 1)] + siz[(rt << 1) + 1];
  if (mx[(rt << 1)] == mx[(rt << 1) + 1]) {
    mx[rt] = mx[(rt << 1)];
    mx2[rt] = max(mx2[(rt << 1)], mx2[(rt << 1) + 1]);
    cnt[rt] = cnt[(rt << 1)] + cnt[(rt << 1) + 1];
  } else if (mx[(rt << 1)] > mx[(rt << 1) + 1]) {
    mx[rt] = mx[(rt << 1)];
    mx2[rt] = max(mx2[(rt << 1)], mx[(rt << 1) + 1]);
    cnt[rt] = cnt[(rt << 1)];
  } else {
    mx[rt] = mx[(rt << 1) + 1];
    mx2[rt] = max(mx2[(rt << 1) + 1], mx[(rt << 1)]);
    cnt[rt] = cnt[(rt << 1) + 1];
  }
}
void Build(int rt, int b, int e) {
  mx[rt] = mx2[rt] = cnt[rt] = siz[rt] = 0;
  sum[rt] = tag1[rt] = tag2[rt] = 0;
  if (b == e) return;
  int mid = (b + e) >> 1;
  Build((rt << 1), b, mid);
  Build((rt << 1) + 1, mid + 1, e);
}
void Insert(int rt, int b, int e, int p, int v) {
  if (b == e) {
    mx[rt] = sum[rt] = v;
    cnt[rt] = siz[rt] = 1;
    return;
  }
  int mid = (b + e) >> 1;
  pushdown(rt);
  if (p <= mid)
    Insert((rt << 1), b, mid, p, v);
  else
    Insert((rt << 1) + 1, mid + 1, e, p, v);
  pushup(rt);
}
void UpdateAdd(int rt, int b, int e, int l, int r, int v) {
  if (l <= b && e <= r) {
    puttag2(rt, v);
    return;
  }
  int mid = (b + e) >> 1;
  pushdown(rt);
  if (l <= mid) UpdateAdd((rt << 1), b, mid, l, r, v);
  if (r > mid) UpdateAdd((rt << 1) + 1, mid + 1, e, l, r, v);
  pushup(rt);
}
void UpdateMin(int rt, int b, int e, int l, int r, int v) {
  if (mx[rt] <= v) return;
  if (l <= b && e <= r && v > mx2[rt]) {
    puttag1(rt, v - mx[rt]);
    return;
  }
  int mid = (b + e) >> 1;
  pushdown(rt);
  if (l <= mid) UpdateMin((rt << 1), b, mid, l, r, v);
  if (r > mid) UpdateMin((rt << 1) + 1, mid + 1, e, l, r, v);
  pushup(rt);
}
int Count(int rt, int b, int e, int r) {
  if (b == e) return siz[rt];
  int mid = (b + e) >> 1;
  if (r <= mid)
    return Count((rt << 1), b, mid, r);
  else
    return siz[(rt << 1)] + Count((rt << 1) + 1, mid + 1, e, r);
}
void Work() {
  Build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    int s = (pos[i] > 1 ? Count(1, 1, n, pos[i] - 1) : 0) + 1;
    if (pos[i] > 1) UpdateMin(1, 1, n, 1, pos[i] - 1, s - 1);
    Insert(1, 1, n, pos[i], i);
    if (pos[i] < n) UpdateAdd(1, 1, n, pos[i] + 1, n, 1);
    ans[i] += sum[1];
  }
}
void init() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    pos[a[i]] = i;
  }
  Work();
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    pos[a[i]] = i;
  }
  Work();
  for (int i = 1; i <= n; ++i) {
    printf("%lld\n", ans[i] - 1LL * i * i);
  }
}
int main() {
  init();
  solve();
  return 0;
}
