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;
inline int input() {
  char ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  int x = 0;
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  return x;
}
inline void output(long long x) {
  static int st[20], top;
  top = 0;
  while (x) st[top++] = x % 10, x /= 10;
  for (; top; --top) putchar(st[top - 1] + '0');
}
int n, op[150010];
int pre[150010], suc[150010];
long long ans[150010];
long long sum;
int tot[150010], ta[150010];
inline void insert(int x, int c) {
  int a = 0;
  for (register int i = x; i; i -= ((i) & -(i))) a += ta[i];
  sum += (long long)a * c;
  for (; x <= n; x += ((x) & -(x))) tot[x] += c;
}
inline void add(int st) {
  for (register int x = st; x <= n; x += ((x) & -(x))) ta[x]++;
  for (register int x = n, y = st - 1; x != y;)
    if (x > y)
      sum += tot[x], x -= ((x) & -(x));
    else
      sum -= tot[y], y -= ((y) & -(y));
}
int fir[150010 * 4], nf[150010 * 4], sec[150010 * 4];
inline void upd(int k) {
  int l = k << 1, r = k << 1 | 1;
  if (fir[l] > fir[r]) {
    fir[k] = fir[l], nf[k] = nf[l];
    sec[k] = ((sec[l]) > (fir[r]) ? (sec[l]) : (fir[r]));
  } else if (fir[l] < fir[r]) {
    fir[k] = fir[r], nf[k] = nf[r];
    sec[k] = ((fir[l]) > (sec[r]) ? (fir[l]) : (sec[r]));
  } else {
    fir[k] = fir[l], nf[k] = nf[l] + nf[r];
    sec[k] = ((sec[l]) > (sec[r]) ? (sec[l]) : (sec[r]));
  }
}
inline void pd(int k) {
  fir[k] < fir[k << 1] ? fir[k << 1] = fir[k] : 0;
  fir[k] < fir[k << 1 | 1] ? fir[k << 1 | 1] = fir[k] : 0;
}
int st, en, c;
void find(int k, int l, int r) {
  if (fir[k] <= c || !nf[k]) return;
  if (c > sec[k]) {
    insert(fir[k], -nf[k]);
    insert(fir[k] = c, nf[k]);
    return;
  }
  if (l == r) return;
  pd(k);
  int mid = l + r >> 1;
  find(k << 1, l, mid);
  find(k << 1 | 1, mid + 1, r);
  upd(k);
}
void getmin(int k, int l, int r) {
  if (fir[k] <= c || !nf[k]) return;
  if (st <= l && r <= en) {
    find(k, l, r);
    return;
  }
  pd(k);
  int mid = l + r >> 1;
  if (st <= mid) getmin(k << 1, l, mid);
  if (mid < en) getmin(k << 1 | 1, mid + 1, r);
  upd(k);
}
void se(int k, int l, int r, int x, int c) {
  if (l == r) {
    if (nf[k])
      insert(fir[k], -1);
    else
      insert(x, -1);
    insert(fir[k] = c, nf[k] = 1);
    return;
  }
  pd(k);
  int mid = l + r >> 1;
  if (x <= mid)
    se(k << 1, l, mid, x, c);
  else
    se(k << 1 | 1, mid + 1, r, x, c);
  upd(k);
}
inline void solve() {
  memset(tot, 0, sizeof(tot)), memset(ta, 0, sizeof(ta));
  memset(fir, 0, sizeof(fir)), memset(nf, 0, sizeof(nf)),
      memset(sec, 0, sizeof(sec));
  sum = 0;
  for (int i = 1; i <= n; ++i) pre[i] = i - 1, suc[i] = i + 1;
  for (int i = n; i >= 1; --i) {
    int x = op[i];
    suc[pre[x]] = suc[x];
    pre[suc[x]] = pre[x];
  }
  int head = n + 1, tail = 0;
  for (int i = 1; i <= n; ++i) {
    int x = op[i];
    head = min(head, x);
    tail = ((tail) > (x) ? (tail) : (x));
    if (head <= pre[x]) st = head, en = pre[x], c = pre[x], getmin(1, 1, n);
    se(1, 1, n, head, tail);
    if (suc[x] <= n) se(1, 1, n, suc[x], tail);
    add(x);
    ans[i] += sum;
  }
}
int main() {
  n = input();
  for (int i = 1; i <= n; ++i) op[input()] = i;
  solve();
  for (int i = 1; i <= n; ++i) op[i] = n - op[i] + 1;
  solve();
  for (int i = 1; i <= n; ++i) output(ans[i] + 1), putchar('\n');
  return 0;
}
