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 maxN = 2e5 + 10, INF = 1e9;
struct Node {
  int mx1, mx2, cnt;
  int t1, t2;
  long long sum;
};
struct Segment {
  Node tree[maxN * 4 + 1];
  void pushup(int node) {
    tree[node].sum = tree[node << 1].sum + tree[node << 1 | 1].sum;
    tree[node].cnt = 0;
    tree[node].mx1 = max(tree[node << 1].mx1, tree[node << 1 | 1].mx1);
    tree[node].mx2 = max(tree[node << 1].mx2, tree[node << 1 | 1].mx2);
    if (tree[node].mx1 != tree[node << 1].mx1)
      tree[node].mx2 = max(tree[node].mx2, tree[node << 1].mx1);
    else
      tree[node].cnt += tree[node << 1].cnt;
    if (tree[node].mx1 != tree[node << 1 | 1].mx1)
      tree[node].mx2 = max(tree[node].mx2, tree[node << 1 | 1].mx1);
    else
      tree[node].cnt += tree[node << 1 | 1].cnt;
  }
  void pushdown(int node, int l, int r) {
    int mid = (l + r) >> 1, t1 = tree[node].t1, t2 = tree[node].t2;
    tree[node].t1 = tree[node].t2 = 0;
    if (t2) {
      tree[node << 1].mx1 += t2;
      tree[node << 1].mx2 += t2;
      tree[node << 1].t2 += t2;
      tree[node << 1].sum += 1ll * (mid - l + 1) * t2;
      tree[node << 1 | 1].mx1 += t2;
      tree[node << 1 | 1].mx2 += t2;
      tree[node << 1 | 1].t2 += t2;
      tree[node << 1 | 1].sum += 1ll * (r - mid) * t2;
    }
    if (t1) {
      int mx = max(tree[node << 1].mx1, tree[node << 1 | 1].mx1);
      if (tree[node << 1].mx1 == mx) {
        tree[node << 1].mx1 += t1;
        tree[node << 1].t1 += t1;
        tree[node << 1].sum += 1ll * tree[node << 1].cnt * t1;
      }
      if (tree[node << 1 | 1].mx1 == mx) {
        tree[node << 1 | 1].mx1 += t1;
        tree[node << 1 | 1].t1 += t1;
        tree[node << 1 | 1].sum += 1ll * tree[node << 1 | 1].cnt * t1;
      }
    }
  }
  void build(int node, int l, int r) {
    tree[node].t1 = tree[node].t2 = 0;
    if (l == r) {
      tree[node].mx1 = tree[node].sum = 0;
      tree[node].cnt = 1;
      tree[node].mx2 = -INF;
      return;
    }
    int mid = (l + r) >> 1;
    build(node << 1, l, mid);
    build(node << 1 | 1, mid + 1, r);
    pushup(node);
  }
  void change(int node, int l, int r, int x, int y, int v) {
    if (x <= l && r <= y) {
      tree[node].mx1 += v;
      tree[node].mx2 += v;
      tree[node].t2 += v;
      tree[node].sum += 1ll * (r - l + 1) * v;
      return;
    }
    pushdown(node, l, r);
    int mid = (l + r) >> 1;
    if (x <= mid) change(node << 1, l, mid, x, y, v);
    if (y > mid) change(node << 1 | 1, mid + 1, r, x, y, v);
    pushup(node);
  }
  void modify(int node, int l, int r, int x, int y, int v) {
    if (tree[node].mx1 <= v) return;
    if (x <= l && r <= y) {
      if (v > tree[node].mx2) {
        tree[node].sum += 1ll * tree[node].cnt * (v - tree[node].mx1);
        tree[node].t1 += v - tree[node].mx1;
        tree[node].mx1 = v;
        return;
      }
      pushdown(node, l, r);
      int mid = (l + r) >> 1;
      modify(node << 1, l, mid, x, y, v);
      modify(node << 1 | 1, mid + 1, r, x, y, v);
      pushup(node);
      return;
    }
    pushdown(node, l, r);
    int mid = (l + r) >> 1;
    if (x <= mid) modify(node << 1, l, mid, x, y, v);
    if (y > mid) modify(node << 1 | 1, mid + 1, r, x, y, v);
    pushup(node);
  }
  long long query(int node, int l, int r, int x, int y) {
    if (x <= l && r <= y) return tree[node].sum;
    pushdown(node, l, r);
    int mid = (l + r) >> 1;
    long long ans = 0;
    if (x <= mid) ans += query(node << 1, l, mid, x, y);
    if (y > mid) ans += query(node << 1 | 1, mid + 1, r, x, y);
    return ans;
  }
} L, R;
int n, a[maxN + 1];
inline int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
  return num * f;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[read()] = i;
  L.build(1, 1, n);
  R.build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    R.change(1, 1, n, x, n, 1);
    L.change(1, 1, n, x, n, -1);
    int v = R.query(1, 1, n, x, x);
    if (x - 1 >= 1) R.modify(1, 1, n, 1, x - 1, v - 1);
    if (x + 1 <= n) L.modify(1, 1, n, x + 1, n, -v);
    L.change(1, 1, n, x, x, v);
    R.change(1, 1, n, x, x, i - v);
    printf("%lld\n", R.query(1, 1, n, 1, n) + L.query(1, 1, n, 1, n));
  }
  return 0;
}
