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 inf = 0x3f3f3f3f;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, p[150055], a[150055];
class Segment_Tree {
 public:
  struct node {
    int mx, se, add1, add2, siz, cmx;
    long long sum;
    node() {
      mx = se = -inf;
      siz = add1 = add2 = cmx = sum = 0;
    }
    inline void Add1(int d) { sum += 1LL * cmx * d, add1 += d, mx += d; }
    inline void Add2(int d) {
      sum += 1LL * siz * d, add2 += d, mx += d, se += d;
    }
  } tree[150055 << 2];
  inline void update(int i) {
    tree[i].sum = tree[i << 1].sum + tree[i << 1 | 1].sum;
    tree[i].siz = tree[i << 1].siz + tree[i << 1 | 1].siz;
    tree[i].mx = max(tree[i << 1].mx, tree[i << 1 | 1].mx);
    tree[i].se = (tree[i << 1].mx == tree[i << 1 | 1].mx
                      ? max(tree[i << 1].se, tree[i << 1 | 1].se)
                      : (tree[i << 1].mx < tree[i << 1 | 1].mx
                             ? max(tree[i << 1].mx, tree[i << 1 | 1].se)
                             : max(tree[i << 1 | 1].mx, tree[i << 1].se)));
    tree[i].cmx =
        (tree[i].mx == tree[i << 1].mx ? tree[i << 1].cmx : 0) +
        (tree[i].mx == tree[i << 1 | 1].mx ? tree[i << 1 | 1].cmx : 0);
  }
  inline void pushdown(int i) {
    if (tree[i].add1) {
      int mx = max(tree[i << 1].mx, tree[i << 1 | 1].mx);
      if (mx == tree[i << 1].mx) tree[i << 1].Add1(tree[i].add1);
      if (mx == tree[i << 1 | 1].mx) tree[i << 1 | 1].Add1(tree[i].add1);
      tree[i].add1 = 0;
    }
    if (tree[i].add2) {
      tree[i << 1].Add2(tree[i].add2);
      tree[i << 1 | 1].Add2(tree[i].add2);
      tree[i].add2 = 0;
    }
  }
  void Min(int i, int L, int R, int l, int r, int d) {
    if (tree[i].mx <= d) return;
    if (L >= l && R <= r && tree[i].se <= d) {
      tree[i].Add1(d - tree[i].mx);
      return;
    }
    pushdown(i);
    int mid = (L + R) >> 1;
    if (l <= mid) Min(i << 1, L, mid, l, r, d);
    if (r > mid) Min(i << 1 | 1, mid + 1, R, l, r, d);
    update(i);
  }
  void Plus(int i, int L, int R, int l, int r, int d) {
    if (L >= l && R <= r) {
      tree[i].Add2(d);
      return;
    }
    pushdown(i);
    int mid = (L + R) >> 1;
    if (l <= mid) Plus(i << 1, L, mid, l, r, d);
    if (r > mid) Plus(i << 1 | 1, mid + 1, R, l, r, d);
    update(i);
  }
  void Change(int i, int L, int R, int x, int d) {
    if (L == R) {
      tree[i].siz = tree[i].cmx = 1;
      tree[i].sum = tree[i].mx = d;
      return;
    }
    pushdown(i);
    int mid = (L + R) >> 1;
    x <= mid ? Change(i << 1, L, mid, x, d)
             : Change(i << 1 | 1, mid + 1, R, x, d);
    update(i);
  }
};
class BIT {
 private:
  int b[150055];
  inline int lowbit(int x) { return x & (-x); }

 public:
  BIT() { memset(b, 0, sizeof(b)); }
  inline void Add(int x, int d) {
    while (x <= n) b[x] += d, x += lowbit(x);
  }
  inline int Ask(int x) {
    int ans = 0;
    while (x) ans += b[x], x -= lowbit(x);
    return ans;
  }
};
long long ans[150055];
void Solve(int opt) {
  Segment_Tree *T = new Segment_Tree;
  BIT *B = new BIT;
  for (int i = 1; i <= n; ++i) {
    p[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    int u = p[i];
    int rk = B->Ask(u) + 1;
    B->Add(u, 1);
    T->Change(1, 1, n, u, i + 1);
    if (u < n) T->Plus(1, 1, n, u + 1, n, 1);
    if (u > 1) T->Min(1, 1, n, 1, u - 1, rk);
    ans[i] +=
        (opt == 1 ? T->tree[1].sum : -(1LL * i * (i + 1) - T->tree[1].sum));
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  Solve(1);
  reverse(a + 1, a + n + 1);
  Solve(-1);
  for (int i = 1; i <= n; ++i) {
    printf("%lld\n", ans[i] - i);
  }
  return 0;
}
