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>
struct node {
  int mx, se, cmx;
  int cnt;
  long long sum;
  node() : mx(0), se(-1), cmx(0), cnt(0), sum(0) {}
  node(int v) : mx(v), se(-1), cmx(1), cnt(1), sum(v) {}
};
node operator+(const node &a, const node &b) {
  node res;
  if (a.mx == b.mx) {
    res.mx = a.mx;
    res.se = std::max(a.se, b.se);
    res.cmx = a.cmx + b.cmx;
  } else if (a.mx > b.mx) {
    res.mx = a.mx;
    res.se = std::max(a.se, b.mx);
    res.cmx = a.cmx;
  } else {
    res.mx = b.mx;
    res.se = std::max(a.mx, b.se);
    res.cmx = b.cmx;
  }
  res.cnt = a.cnt + b.cnt;
  res.sum = a.sum + b.sum;
  return res;
}
int enlarge(int n) {
  int res = 1;
  while (res < n) {
    res <<= 1;
  }
  return res;
}
class segment_tree {
  int n;
  std::vector<node> a;
  std::vector<int> tag, mxtag;
  void node_add(int u, int v) {
    a[u].mx += v;
    a[u].se += v;
    a[u].sum += 1ll * v * a[u].cnt;
    tag[u] += v;
  }
  void node_add_mx(int u, int v) {
    a[u].mx += v;
    a[u].sum += 1ll * v * a[u].cmx;
    mxtag[u] += v;
  }
  void down(int u) {
    if (tag[u]) {
      node_add(u << 1, tag[u]);
      node_add(u << 1 | 1, tag[u]);
      tag[u] = 0;
    }
    if (mxtag[u]) {
      int lmx = a[u << 1].mx, rmx = a[u << 1 | 1].mx;
      if (lmx >= rmx) {
        node_add_mx(u << 1, mxtag[u]);
      }
      if (rmx >= lmx) {
        node_add_mx(u << 1 | 1, mxtag[u]);
      }
      mxtag[u] = 0;
    }
  }
  void modify(int u, int l, int r, int x, int v) {
    if (l + 1 == r) {
      a[u] = node(v);
      return;
    }
    int mid = (l + r + 1) >> 1;
    down(u);
    if (x < mid) {
      modify(u << 1, l, mid, x, v);
    } else {
      modify(u << 1 | 1, mid, r, x, v);
    }
    a[u] = a[u << 1] + a[u << 1 | 1];
  }
  void add(int u, int l, int r, int L, int R, int v) {
    if (L <= l && r <= R) {
      node_add(u, v);
      return;
    }
    int mid = (l + r + 1) >> 1;
    down(u);
    if (L < mid) {
      add(u << 1, l, mid, L, R, v);
    }
    if (mid < R) {
      add(u << 1 | 1, mid, r, L, R, v);
    }
    a[u] = a[u << 1] + a[u << 1 | 1];
  }
  void chmin(int u, int l, int r, int L, int R, int v) {
    if (a[u].mx <= v) {
      return;
    }
    if (L <= l && r <= R && a[u].se < v) {
      node_add_mx(u, v - a[u].mx);
      return;
    }
    int mid = (l + r + 1) >> 1;
    down(u);
    if (L < mid) {
      chmin(u << 1, l, mid, L, R, v);
    }
    if (mid < R) {
      chmin(u << 1 | 1, mid, r, L, R, v);
    }
    a[u] = a[u << 1] + a[u << 1 | 1];
  }
  int get_cnt(int u, int l, int r, int L, int R) {
    if (L <= l && r <= R) {
      return a[u].cnt;
    }
    int mid = (l + r + 1) >> 1;
    down(u);
    if (R <= mid) {
      return get_cnt(u << 1, l, mid, L, R);
    } else if (L >= mid) {
      return get_cnt(u << 1 | 1, mid, r, L, R);
    } else {
      return get_cnt(u << 1, l, mid, L, R) + get_cnt(u << 1 | 1, mid, r, L, R);
    }
  }

 public:
  segment_tree(int _n)
      : n(_n),
        a(enlarge(n) << 1),
        tag(enlarge(n) << 1),
        mxtag(enlarge(n) << 1) {}
  void modify(int x, int v) { modify(1, 0, n, x, v); }
  void add(int l, int r, int v) {
    if (l >= r) {
      return;
    }
    add(1, 0, n, l, r, v);
  }
  void chmin(int l, int r, int v) {
    if (l >= r) {
      return;
    }
    chmin(1, 0, n, l, r, v);
  }
  int get_cnt(int l, int r) {
    if (l >= r) {
      return 0;
    }
    return get_cnt(1, 0, n, l, r);
  }
  long long get_sum() { return a[1].sum; }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int n;
  std::cin >> n;
  std::vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    int v;
    std::cin >> v;
    --v;
    p[v] = i;
  }
  std::vector<long long> ans(n);
  for (int _ = 0; _ < 2; ++_) {
    segment_tree T(n);
    for (int i = 0; i < n; ++i) {
      T.modify(p[i], i + 1);
      T.chmin(0, p[i], T.get_cnt(0, p[i]));
      T.add(p[i] + 1, n, 1);
      ans[i] += T.get_sum() - 1ll * i * (i + 1) / 2;
    }
    for (int i = 0; i < n; ++i) {
      p[i] = n - 1 - p[i];
    }
  }
  for (int i = 0; i < n; ++i) {
    std::cout << ans[i] - i - 1 << "\n";
  }
}
