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;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int x = 0, p = 1;
  char c = IO::_getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = IO::_getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = IO::_getchar();
  return x * p;
}
const int Maxn = 150005;
long long bit[Maxn];
void Add(int pos) {
  while (pos < Maxn) {
    bit[pos]++;
    pos += pos & -pos;
  }
}
long long Query(int pos) {
  long long ret = 0;
  while (pos) {
    ret += bit[pos];
    pos -= pos & -pos;
  }
  return ret;
}
long long Ans;
struct node {
  int Mx1, Mx2, cnt, tag;
  long long Sum;
  node() {
    Mx1 = Mx2 = 0;
    cnt = -1;
    Sum = 0;
    tag = -1145141919;
  }
  node(int Val) {
    Mx1 = Val;
    Mx2 = -1e9;
    cnt = 1;
    tag = -1145141919;
    Sum = Val;
  }
} Ept;
struct AngelBeats {
  node tree[Maxn * 4];
  void Set(int p, int val, int t = 0) {
    if (t && val > 0) {
      long long dt = Query(tree[p].Mx1 - 1) - Query(val - 1);
      Ans -= dt * tree[p].cnt;
    }
    if (t && val < 0) {
      long long dt = Query(-val) - Query(-tree[p].Mx1);
      Ans -= dt * tree[p].cnt;
    }
    long long dlt = val - tree[p].Mx1;
    tree[p].Sum += dlt * tree[p].cnt;
    tree[p].Mx1 = val;
  }
  void push_down(int p) {
    if (tree[p].tag != -1145141919) {
      if (tree[p * 2].cnt != -1 && tree[p * 2].Mx1 > tree[p].tag) {
        Set(p * 2, tree[p].tag);
        tree[p * 2].tag = tree[p].tag;
      }
      if (tree[p * 2 + 1].cnt != -1 && tree[p * 2 + 1].Mx1 > tree[p].tag) {
        Set(p * 2 + 1, tree[p].tag);
        tree[p * 2 + 1].tag = tree[p].tag;
      }
      tree[p].tag = -1145141919;
    }
  }
  void push_up(int p) {
    if (tree[p * 2].cnt == -1) {
      tree[p] = tree[p * 2 + 1];
      return;
    }
    if (tree[p * 2 + 1].cnt == -1) {
      tree[p] = tree[p * 2];
      return;
    }
    if (tree[p * 2].Mx1 == tree[p * 2 + 1].Mx1) {
      tree[p].Mx1 = tree[p * 2].Mx1;
      tree[p].cnt = tree[p * 2].cnt + tree[p * 2 + 1].cnt;
      tree[p].Mx2 = max(tree[p * 2].Mx2, tree[p * 2 + 1].Mx2);
    } else if (tree[p * 2].Mx1 < tree[p * 2 + 1].Mx1) {
      tree[p].Mx1 = tree[p * 2 + 1].Mx1;
      tree[p].cnt = tree[p * 2 + 1].cnt;
      tree[p].Mx2 = max(tree[p * 2 + 1].Mx2, tree[p * 2].Mx1);
    } else {
      tree[p].Mx1 = tree[p * 2].Mx1;
      tree[p].cnt = tree[p * 2].cnt;
      tree[p].Mx2 = max(tree[p * 2].Mx2, tree[p * 2 + 1].Mx1);
    }
    tree[p].Sum = tree[p * 2].Sum + tree[p * 2 + 1].Sum;
  }
  void SetVal(int p, int l, int r, int pos, int val) {
    if (l == r) {
      tree[p] = node(val);
      return;
    }
    push_down(p);
    int mid = l + r >> 1;
    if (pos <= mid) {
      SetVal(p * 2, l, mid, pos, val);
    } else {
      SetVal(p * 2 + 1, mid + 1, r, pos, val);
    }
    push_up(p);
  }
  void modify(int p, int l, int r, int lo, int hi, int val) {
    if (lo > hi) return;
    if (tree[p].cnt == -1 || val >= tree[p].Mx1) return;
    if (lo <= l && r <= hi) {
      if (val > tree[p].Mx2) {
        Set(p, val, 1);
        tree[p].tag = val;
        return;
      }
    }
    if (l == r) return;
    push_down(p);
    int mid = l + r >> 1;
    if (lo <= mid) {
      modify(p * 2, l, mid, lo, min(hi, mid), val);
    }
    if (hi > mid) {
      modify(p * 2 + 1, mid + 1, r, max(lo, mid + 1), hi, val);
    }
    push_up(p);
  }
} L, R;
int n, A[Maxn];
int pos[Maxn];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) A[i] = read();
  for (int i = 1; i <= n; i++) pos[A[i]] = i;
  Ans = 0;
  for (int i = 1; i <= n; i++) {
    int p = pos[i];
    R.SetVal(1, 1, n, p, n + 1);
    L.SetVal(1, 1, n, p, 0);
    Ans += i;
    R.modify(1, 1, n, 1, p - 1, p);
    L.modify(1, 1, n, p + 1, n, -p);
    Add(p);
    printf("%lld\n", Ans);
  }
  return ~~(0 ^ 0 ^ 0);
}
