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;
int n, a[150100], p[150100];
long long res[150100];
struct SegBeat {
  int mx, mmx, mxc, tag, gat, sz;
  long long sum;
  SegBeat() { sum = tag = -1, sz = gat = 0; }
} seg[600100];
void pushup(int x) {
  if (seg[x << 1].sum == -1 && seg[x << 1 | 1].sum == -1) {
    seg[x].sum = -1;
    return;
  }
  if (seg[x << 1].sum == -1) {
    seg[x] = seg[x << 1 | 1], seg[x].gat = 0, seg[x].tag = -1;
    return;
  }
  if (seg[x << 1 | 1].sum == -1) {
    seg[x] = seg[x << 1], seg[x].gat = 0, seg[x].tag = -1;
    return;
  }
  seg[x].mx = max(seg[x << 1].mx, seg[x << 1 | 1].mx), seg[x].mmx = -1,
  seg[x].mxc = 0, seg[x].sum = seg[x << 1].sum + seg[x << 1 | 1].sum,
  seg[x].sz = seg[x << 1].sz + seg[x << 1 | 1].sz;
  if (seg[x].mx != seg[x << 1].mx)
    seg[x].mmx = max(seg[x].mmx, seg[x << 1].mx);
  else
    seg[x].mxc += seg[x << 1].mxc,
        seg[x].mmx = max(seg[x].mmx, seg[x << 1].mmx);
  if (seg[x].mx != seg[x << 1 | 1].mx)
    seg[x].mmx = max(seg[x].mmx, seg[x << 1 | 1].mx);
  else
    seg[x].mxc += seg[x << 1 | 1].mxc,
        seg[x].mmx = max(seg[x].mmx, seg[x << 1 | 1].mmx);
}
void MOD(int x, int y) {
  seg[x].sum -= 1ll * (seg[x].mx - y) * seg[x].mxc, seg[x].mx = seg[x].tag = y;
}
void ADD(int x, int y) {
  seg[x].mx += y, seg[x].gat += y, seg[x].mmx += y,
      seg[x].sum += 1ll * seg[x].sz * y;
  if (seg[x].tag != -1) seg[x].tag += y;
}
void pushdown(int x) {
  if (seg[x << 1].sum != -1) ADD(x << 1, seg[x].gat);
  if (seg[x << 1 | 1].sum != -1) ADD(x << 1 | 1, seg[x].gat);
  seg[x].gat = 0;
  if (seg[x].tag == -1) return;
  if (seg[x << 1].sum != -1 && seg[x << 1].mx > seg[x].tag)
    MOD(x << 1, seg[x].tag);
  if (seg[x << 1 | 1].sum != -1 && seg[x << 1 | 1].mx > seg[x].tag)
    MOD(x << 1 | 1, seg[x].tag);
  seg[x].tag = -1;
}
void modifypos(int x, int l, int r, int P, int val) {
  if (l > P || r < P) return;
  if (l == r) {
    seg[x].mx = val, seg[x].mmx = -1, seg[x].sum = val, seg[x].mxc = 1,
    seg[x].sz = 1;
    return;
  }
  pushdown(x), modifypos(x << 1, l, ((l + r) >> 1), P, val),
      modifypos(x << 1 | 1, ((l + r) >> 1) + 1, r, P, val), pushup(x);
}
void modifymn(int x, int l, int r, int L, int R, int val) {
  if (seg[x].mx <= val || seg[x].sum == -1 || l > R || r < L) return;
  if (L <= l && r <= R && seg[x].mmx < val) {
    MOD(x, val);
    return;
  }
  pushdown(x), modifymn(x << 1, l, ((l + r) >> 1), L, R, val),
      modifymn(x << 1 | 1, ((l + r) >> 1) + 1, r, L, R, val), pushup(x);
}
void modifyshift(int x, int l, int r, int L, int R) {
  if (l > R || r < L || seg[x].sum == -1) return;
  if (L <= l && r <= R) {
    ADD(x, 1);
    return;
  }
  pushdown(x), modifyshift(x << 1, l, ((l + r) >> 1), L, R),
      modifyshift(x << 1 | 1, ((l + r) >> 1) + 1, r, L, R), pushup(x);
}
void clear(int x, int l, int r) {
  seg[x] = SegBeat();
  if (l != r)
    clear(x << 1, l, ((l + r) >> 1)), clear(x << 1 | 1, ((l + r) >> 1) + 1, r);
}
int t[200100];
void BITADD(int x) {
  while (x <= n) t[x]++, x += x & -x;
}
int BITSUM(int x) {
  int ret = 0;
  while (x) ret += t[x], x -= x & -x;
  return ret;
}
void func() {
  clear(1, 1, n), memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) {
    int Rmax = BITSUM(p[i]);
    modifymn(1, 1, n, 1, p[i] - 1, Rmax);
    modifypos(1, 1, n, p[i], i);
    modifyshift(1, 1, n, p[i] + 1, n);
    res[i] += seg[1].sum;
    BITADD(p[i]);
  }
}
inline void read(int &x) {
  x = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
}
inline void print(long long x) {
  if (x <= 9)
    putchar('0' + x);
  else
    print(x / 10), putchar('0' + x % 10);
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]), p[a[i]] = i;
  func();
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  func();
  for (int i = 1; i <= n; i++) print(res[i] - 1ll * i * i), putchar('\n');
  return 0;
}
