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>
const int N = 1.5e5 + 5;
int n, a[N], p[N];
long long ans[N];
struct node {
  int v1, v2, cv, cnt;
  int t1, t2;
  long long sum;
  inline void tag1(int v) {
    v1 += v;
    t1 += v;
    sum += 1ll * v * cv;
  }
  inline void tag2(int v) {
    v1 += v;
    v2 += v;
    t2 += v;
    sum += 1ll * v * cnt;
  }
} t[N << 2];
int L, R, pos, v;
inline void pushup(int p) {
  int x = p << 1, y = p << 1 | 1;
  if (t[x].v1 == t[y].v1) {
    t[p].v1 = t[x].v1;
    t[p].cv = t[x].cv + t[y].cv;
    t[p].v2 = std::max(t[x].v2, t[y].v2);
  } else {
    if (t[x].v1 < t[y].v1) {
      std::swap(x, y);
    }
    t[p].v1 = t[x].v1;
    t[p].cv = t[x].cv;
    t[p].v2 = std::max(t[x].v2, t[y].v1);
  }
  t[p].sum = t[p << 1].sum + t[p << 1 | 1].sum;
  t[p].cnt = t[p << 1].cnt + t[p << 1 | 1].cnt;
}
void build(int p = 1, int l = 1, int r = n) {
  t[p].t1 = t[p].t2 = t[p].cnt = t[p].cv = 0;
  t[p].v1 = t[p].v2 = t[p].sum = 0;
  if (l != r) {
    int mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
  }
}
inline void pushdown(int p) {
  int l = p << 1, r = p << 1 | 1;
  if (t[p].t1) {
    int max = std::max(t[l].v1, t[r].v1);
    if (t[l].v1 == max) {
      t[l].tag1(t[p].t1);
    }
    if (t[r].v1 == max) {
      t[r].tag1(t[p].t1);
    }
    t[p].t1 = 0;
  }
  if (t[p].t2) {
    t[l].tag2(t[p].t2);
    t[r].tag2(t[p].t2);
    t[p].t2 = 0;
  }
}
void change(int p = 1, int l = 1, int r = n) {
  if (t[p].v1 <= v) {
    return;
  }
  if (l >= L && r <= R && t[p].v2 <= v) {
    t[p].tag1(v - t[p].v1);
    return;
  }
  pushdown(p);
  int mid = l + r >> 1;
  if (mid >= L) {
    change(p << 1, l, mid);
  }
  if (mid < R) {
    change(p << 1 | 1, mid + 1, r);
  }
  pushup(p);
}
void add(int p = 1, int l = 1, int r = n) {
  if (l >= L && r <= R) {
    t[p].tag2(1);
    return;
  }
  pushdown(p);
  int mid = l + r >> 1;
  if (mid >= L) {
    add(p << 1, l, mid);
  }
  if (mid < R) {
    add(p << 1 | 1, mid + 1, r);
  }
  pushup(p);
}
int query(int p = 1, int l = 1, int r = n) {
  if (l >= L && r <= R) {
    return t[p].cnt;
  }
  int mid = l + r >> 1;
  if (mid >= L && mid < R) {
    return query(p << 1, l, mid) + query(p << 1 | 1, mid + 1, r);
  } else if (mid >= L) {
    return query(p << 1, l, mid);
  } else {
    return query(p << 1 | 1, mid + 1, r);
  }
}
void activate(int p = 1, int l = 1, int r = n) {
  if (l == r) {
    t[p].cnt = t[p].cv = 1;
    t[p].v1 = t[p].sum = v;
    return;
  }
  pushdown(p);
  int mid = l + r >> 1;
  if (mid >= pos) {
    activate(p << 1, l, mid);
  } else {
    activate(p << 1 | 1, mid + 1, r);
  }
  pushup(p);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  auto process = [&]() {
    build();
    for (int i = 1; i <= n; ++i) {
      p[a[i]] = i;
    }
    for (int i = 1; i <= n; ++i) {
      int pos = p[i];
      L = 1, R = pos;
      v = query();
      change();
      L = pos, R = n;
      add();
      ::pos = pos, v = i;
      activate();
      ans[i] += t[1].sum;
    }
  };
  process();
  std::reverse(a + 1, a + n + 1);
  process();
  for (int i = 1; i <= n; ++i) {
    printf("%lld\n", ans[i] - 1ll * i * i);
  }
  return 0;
}
