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>
constexpr int SIZE = 300;
int n, cnt;
std::vector<int> a, pos, ltag, rtag, l, r, lp, rp, atag, temp1, temp2;
std::vector<std::vector<int>> sl, sr;
long long ans;
void rebuild(int b) {
  for (int i = b * SIZE; i < std::min(n, (b + 1) * SIZE); ++i) {
    if (l[i] == r[i]) continue;
    l[i] = std::max(l[i] + atag[b], ltag[b]);
    r[i] = std::min(r[i] + atag[b], rtag[b]);
  }
  ltag[b] = -1;
  rtag[b] = n;
  atag[b] = 0;
  lp[b] = rp[b] = sr[b].size() + 1;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n;
  a.resize(n);
  pos.resize(n);
  cnt = (n + SIZE - 1) / SIZE;
  l.resize(n);
  r.resize(n);
  ltag.assign(cnt, -1);
  rtag.assign(cnt, n);
  atag.resize(cnt);
  sl.resize(cnt);
  sr.resize(cnt);
  lp.resize(cnt);
  rp.resize(cnt);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    --a[i];
    pos[a[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    int b = pos[i] / SIZE;
    rebuild(b);
    int rpos = 0;
    for (int j = 0; j < b; ++j) rpos += sl[j].size();
    for (int j = b * SIZE; j < pos[i]; ++j)
      if (l[j] != r[j]) ++rpos;
    l[pos[i]] = -1;
    r[pos[i]] = i + 1;
    ans += i + 1;
    for (int j = b * SIZE; j < pos[i]; ++j) {
      if (l[j] == r[j]) continue;
      if (r[j] > rpos) {
        ans -= r[j] - rpos;
        r[j] = rpos;
      }
    }
    for (int j = pos[i] + 1; j < std::min(n, (b + 1) * SIZE); ++j) {
      if (l[j] == r[j]) continue;
      ++l[j];
      ++r[j];
      if (l[j] < rpos) {
        ans -= rpos - l[j];
        l[j] = rpos;
      }
    }
    temp1.clear();
    temp2.clear();
    for (int j : sr[b]) {
      if (j < pos[i]) {
        temp1.push_back(j);
      } else {
        temp2.push_back(j);
      }
    }
    std::merge(temp1.begin(), temp1.end(), temp2.begin(), temp2.end(),
               sr[b].begin(), [&](int i, int j) { return r[i] < r[j]; });
    temp1.clear();
    temp2.clear();
    for (int j : sl[b]) {
      if (j < pos[i]) {
        temp1.push_back(j);
      } else {
        temp2.push_back(j);
      }
    }
    std::merge(temp1.begin(), temp1.end(), temp2.begin(), temp2.end(),
               sl[b].begin(), [&](int i, int j) { return l[i] > l[j]; });
    sl[b].push_back(pos[i]);
    sr[b].push_back(pos[i]);
    for (int j = 0; j < b; ++j) {
      if (rpos >= rtag[j]) continue;
      ans -= 1LL * (sr[j].size() - rp[j]) * (rtag[j] - rpos);
      while (rp[j] > 0 && r[sr[j][rp[j] - 1]] + atag[j] > rpos) {
        ans -= r[sr[j][rp[j] - 1]] + atag[j] - rpos;
        --rp[j];
      }
      rtag[j] = rpos;
    }
    for (int j = b + 1; j < cnt; ++j) {
      ++atag[j];
      ++ltag[j];
      ++rtag[j];
      if (rpos <= ltag[j]) continue;
      ans -= 1LL * (sl[j].size() - lp[j]) * (rpos - ltag[j]);
      while (lp[j] > 0 && l[sl[j][lp[j] - 1]] + atag[j] < rpos) {
        ans -= rpos - (l[sl[j][lp[j] - 1]] + atag[j]);
        --lp[j];
      }
      ltag[j] = rpos;
    }
    std::cout << ans << "\n";
  }
  return 0;
}
