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>
const int maxn = 2e5 + 5;
const int inf = 1e9;
using namespace std;
int rd() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
struct node {
  int mx, cmx, cnt, s;
  long long sum;
  node() { mx = cmx = cnt = s = sum = 0; }
  node(int _mx, int _cmx, int _cnt, int _s, long long _sum) {
    mx = _mx;
    cmx = _cmx;
    cnt = _cnt;
    s = _s;
    sum = _sum;
  }
  node operator+(const node &a) const {
    node c;
    c.sum = sum + a.sum;
    c.s = s + a.s;
    if (mx == a.mx) {
      c.mx = mx;
      c.cnt = cnt + a.cnt;
      c.cmx = max(cmx, a.cmx);
    } else if (mx > a.mx) {
      c.mx = mx;
      c.cnt = cnt;
      c.cmx = max(a.mx, cmx);
    } else {
      c.mx = a.mx;
      c.cnt = a.cnt;
      c.cmx = max(a.cmx, mx);
    }
    return c;
  }
};
struct Segmemt_tree {
  node sgt[maxn << 2];
  int tag[maxn << 2], num[maxn << 2];
  void build(int p, int l, int r) {
    sgt[p] = node(0, -inf, 0, 0, 0LL);
    tag[p] = 0;
    num[p] = inf;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build((p << 1), l, mid);
    build((p << 1 | 1), mid + 1, r);
  }
  void add(int p, int v) {
    if (!sgt[p].s) return;
    sgt[p].mx += v;
    sgt[p].sum += 1LL * v * sgt[p].s;
    if (sgt[p].cmx != -inf) sgt[p].cmx += v;
    tag[p] += v;
    if (num[p] != inf) num[p] += v;
  }
  void change(int p, int v) {
    if (!sgt[p].s) return;
    if (v >= sgt[p].mx) return;
    sgt[p].sum -= 1LL * sgt[p].cnt * (sgt[p].mx - v);
    sgt[p].mx = v;
    num[p] = v;
  }
  void pushdown(int p) {
    if (tag[p]) {
      add((p << 1), tag[p]);
      add((p << 1 | 1), tag[p]);
      tag[p] = 0;
    }
    if (num[p] != inf) {
      change((p << 1), num[p]);
      change((p << 1 | 1), num[p]);
      num[p] = inf;
    }
  }
  void modify(int p, int l, int r, int sl, int sr, int v) {
    if (!sgt[p].s) return;
    if (sl <= l && r <= sr) {
      add(p, v);
      return;
    }
    int mid = (l + r) >> 1;
    pushdown(p);
    if (sl <= mid) modify((p << 1), l, mid, sl, sr, v);
    if (mid < sr) modify((p << 1 | 1), mid + 1, r, sl, sr, v);
    sgt[p] = sgt[(p << 1)] + sgt[(p << 1 | 1)];
  }
  void insert(int p, int l, int r, int x, int z) {
    if (l == r) {
      sgt[p] = (node){z, -inf, 1, 1, z};
      return;
    }
    int mid = (l + r) >> 1;
    pushdown(p);
    if (x <= mid)
      insert((p << 1), l, mid, x, z);
    else
      insert((p << 1 | 1), mid + 1, r, x, z);
    sgt[p] = sgt[(p << 1)] + sgt[(p << 1 | 1)];
  }
  void getmin(int p, int l, int r, int sl, int sr, int v) {
    if (!sgt[p].s) return;
    if (sl <= l && r <= sr) {
      if (v > sgt[p].cmx) {
        change(p, v);
        return;
      }
    }
    if (l == r) return;
    int mid = (l + r) >> 1;
    pushdown(p);
    if (sl <= mid) getmin((p << 1), l, mid, sl, sr, v);
    if (mid < sr) getmin((p << 1 | 1), mid + 1, r, sl, sr, v);
    sgt[p] = sgt[(p << 1)] + sgt[(p << 1 | 1)];
  }
  int query(int p, int l, int r, int sl, int sr) {
    if (sl <= l && r <= sr) return sgt[p].s;
    int mid = (l + r) >> 1, res = 0;
    pushdown(p);
    if (sl <= mid) res = query((p << 1), l, mid, sl, sr);
    if (mid < sr) res += query((p << 1 | 1), mid + 1, r, sl, sr);
    return res;
  }
} T;
int n, a[maxn], pos[maxn];
long long ans[maxn];
void solve() {
  T.build(1, 1, n);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    if (pos[i] + 1 <= n) T.modify(1, 1, n, pos[i] + 1, n, 1);
    if (pos[i] - 1 >= 1) {
      int x = T.query(1, 1, n, 1, pos[i] - 1) + 1;
      T.getmin(1, 1, n, 1, pos[i] - 1, x);
    }
    T.insert(1, 1, n, pos[i], i + 1);
    ans[i] += T.sgt[1].sum;
  }
}
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) a[i] = rd();
  solve();
  reverse(a + 1, a + n + 1);
  solve();
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] - 1LL * i * (i + 2));
  return 0;
}
