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;
class tree {
 public:
  long long dat[150010];
  int N;
  void init(int tN) {
    N = tN;
    memset(dat, 0, sizeof(dat));
  }
  inline int lowbit(int x) { return x & (-x); }
  inline void modify(int p, long long v) {
    if (!p) return;
    for (; p <= N; p += lowbit(p)) dat[p] += v;
  }
  inline long long query(int p) {
    long long ret = 0;
    for (; p; p -= lowbit(p)) ret += dat[p];
    return ret;
  }
} T1, T2;
int n, a[150010], pos[150010];
long long ans[150010];
class node {
 public:
  int mx, mxc, sec;
} t[600010];
vector<pair<int, long long> > opt;
void pushup(int p) {
  t[p].mx = max(t[(p << 1)].mx, t[((p << 1) | 1)].mx);
  t[p].mxc = 0;
  if (t[(p << 1)].mx == t[p].mx) t[p].mxc += t[(p << 1)].mxc;
  if (t[((p << 1) | 1)].mx == t[p].mx) t[p].mxc += t[((p << 1) | 1)].mxc;
  t[p].sec = max(t[(p << 1)].sec, t[((p << 1) | 1)].sec);
  if (t[(p << 1)].mx != t[p].mx) t[p].sec = max(t[p].sec, t[(p << 1)].mx);
  if (t[((p << 1) | 1)].mx != t[p].mx)
    t[p].sec = max(t[p].sec, t[((p << 1) | 1)].mx);
}
void pushnow(int p, int v) {
  if (t[p].mx > v) t[p].mx = v;
}
void pushdown(int p) {
  pushnow((p << 1), t[p].mx);
  pushnow(((p << 1) | 1), t[p].mx);
}
void modify(int p, int l, int r, int ml, int mr, int v) {
  if (v >= t[p].mx) return;
  if (ml <= l && r <= mr && t[p].sec < v) {
    opt.push_back(make_pair(t[p].mx, -t[p].mxc));
    t[p].mx = v;
    opt.push_back(make_pair(t[p].mx, t[p].mxc));
    return;
  }
  pushdown(p);
  int mid = (l + r) >> 1;
  if (ml <= mid) modify((p << 1), l, mid, ml, mr, v);
  if (mr > mid) modify(((p << 1) | 1), mid + 1, r, ml, mr, v);
  pushup(p);
}
void change(int p, int l, int r, int pos, int v) {
  if (l == r) {
    opt.push_back(make_pair(t[p].mx, -t[p].mxc));
    t[p].mx = v;
    opt.push_back(make_pair(t[p].mx, t[p].mxc));
    return;
  }
  pushdown(p);
  int mid = (l + r) >> 1;
  if (pos <= mid)
    change((p << 1), l, mid, pos, v);
  else
    change(((p << 1) | 1), mid + 1, r, pos, v);
  pushup(p);
}
void build(int p, int l, int r) {
  t[p].mx = 0;
  t[p].mxc = r - l + 1;
  t[p].sec = -1e9;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build((p << 1), l, mid);
  build(((p << 1) | 1), mid + 1, r);
}
void solve(bool type) {
  T1.init(n);
  T2.init(n);
  build(1, 1, n);
  long long tot = 0;
  set<int> s1, s2;
  s1.insert(pos[1]);
  s2.insert(pos[1]);
  T1.modify(pos[1], 1);
  int mx = pos[1];
  change(1, 1, n, pos[1], pos[1]);
  T2.modify(pos[1], -1);
  for (int i = 2; i <= n; i++) {
    int cur = pos[i];
    auto it = s1.lower_bound(cur);
    int nxt = 0;
    if (it != s1.end()) nxt = *it;
    if (it == s1.begin()) {
      tot -= T1.query(cur);
      T2.modify(cur, -1);
      s2.insert(cur);
      change(1, 1, n, cur, mx);
    } else {
      auto tmp = it;
      --tmp;
      int pre = *tmp;
      modify(1, 1, n, 1, pre, pre);
      if (it == s1.end())
        mx = cur;
      else {
        change(1, 1, n, nxt, mx);
        if (s2.find(nxt) == s2.end()) {
          tot -= T1.query(nxt);
          T2.modify(nxt, -1);
          s2.insert(nxt);
        }
      }
      change(1, 1, n, *s1.begin(), mx);
    }
    s1.insert(cur);
    tot += T2.query(n) - T2.query(cur - 1);
    T1.modify(cur, 1);
    while (!opt.empty()) {
      tot += opt.back().second * T1.query(opt.back().first);
      T2.modify(opt.back().first, opt.back().second);
      opt.pop_back();
    }
    ans[i] += tot;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  solve(0);
  for (int i = 1; i <= n; i++) pos[i] = n + 1 - pos[i];
  solve(1);
  ans[1] += 2;
  for (int i = 1; i <= n; i++) printf("%lld\n", ans[i] - 1);
  return 0;
}
