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 N = 150005, INF = 1E9 + 7;
int n, u, a[N];
enum operation { CHMIN, CHMAX, ADD };
struct node {
  int mi, mic, mis;
  int mx, mxc, mxs;
  int lz;
  long long sum;
  node(int _n = 0) {
    mi = mx = _n;
    mic = mxc = 1;
    mis = INF;
    mxs = -INF;
    lz = 0;
    sum = _n;
  }
  node operator+(const node &oth) const {
    node ans;
    ans.sum = sum + oth.sum;
    ans.mi = min(mi, oth.mi);
    if (mi == oth.mi) {
      ans.mic = mic + oth.mic;
      ans.mis = min(mis, oth.mis);
    } else if (mi < oth.mi) {
      ans.mic = mic;
      ans.mis = min(mis, oth.mi);
    } else {
      ans.mic = oth.mic;
      ans.mis = min(mi, oth.mis);
    }
    ans.mx = max(mx, oth.mx);
    if (mx == oth.mx) {
      ans.mxc = mxc + oth.mxc;
      ans.mxs = max(mxs, oth.mxs);
    } else if (mx > oth.mx) {
      ans.mxc = mxc;
      ans.mxs = max(mxs, oth.mx);
    } else {
      ans.mxc = oth.mxc;
      ans.mxs = max(mx, oth.mxs);
    }
    return ans;
  }
};
struct segment_tree {
  node tr[4 * N];
  bool break_condition(int i, int v, operation op) {
    switch (op) {
      case CHMIN:
        return tr[i].mx <= v;
      case CHMAX:
        return tr[i].mi >= v;
      case ADD:
        return false;
    }
  }
  bool tag_condition(int i, int v, operation op) {
    switch (op) {
      case CHMIN:
        return tr[i].mxs < v;
      case CHMAX:
        return tr[i].mis > v;
      case ADD:
        return true;
    }
  }
  void tag(int l, int r, int i, int v, operation op) {
    switch (op) {
      case CHMIN: {
        tr[i].sum -= 1LL * (tr[i].mx - v) * tr[i].mxc;
        if (tr[i].mi == tr[i].mx) {
          tr[i].mi = v;
        } else if (tr[i].mis == tr[i].mx) {
          tr[i].mis = v;
        }
        tr[i].mx = v;
        return;
      }
      case CHMAX: {
        tr[i].sum += 1LL * (v - tr[i].mi) * tr[i].mic;
        if (tr[i].mx == tr[i].mi) {
          tr[i].mx = v;
        } else if (tr[i].mxs == tr[i].mi) {
          tr[i].mxs = v;
        }
        tr[i].mi = v;
        return;
      }
      case ADD: {
        tr[i].sum += 1LL * (r - l + 1) * v;
        tr[i].lz += v;
        tr[i].mi += v;
        tr[i].mis += v;
        tr[i].mx += v;
        tr[i].mxs += v;
        return;
      }
    }
  }
  void push(int l, int r, int i) {
    tag(l, (l + r) / 2, i * 2, tr[i].lz, ADD);
    tag((l + r) / 2 + 1, r, i * 2 + 1, tr[i].lz, ADD);
    tr[i].lz = 0;
    for (operation op : {CHMIN, CHMAX}) {
      int v = (op == CHMIN ? tr[i].mx : tr[i].mi);
      if (!break_condition(i * 2, v, op)) {
        tag(l, (l + r) / 2, i * 2, v, op);
      }
      if (!break_condition(i * 2 + 1, v, op)) {
        tag((l + r) / 2 + 1, r, i * 2 + 1, v, op);
      }
    }
  }
  void build(int l, int r, int i, int v) {
    if (l == r) {
      tr[i] = node(v);
    } else {
      build(l, (l + r) / 2, i * 2, v);
      build((l + r) / 2 + 1, r, i * 2 + 1, v);
      tr[i] = tr[i * 2] + tr[i * 2 + 1];
    }
  }
  void update(int l, int r, int i, int L, int R, int v, operation op) {
    if (l > R || r < L || break_condition(i, v, op)) {
      return;
    } else if (L <= l && r <= R && tag_condition(i, v, op)) {
      tag(l, r, i, v, op);
    } else {
      push(l, r, i);
      update(l, (l + r) / 2, i * 2, L, R, v, op);
      update((l + r) / 2 + 1, r, i * 2 + 1, L, R, v, op);
      tr[i] = tr[i * 2] + tr[i * 2 + 1];
    }
  }
  long long query(int l, int r, int i, int L, int R) {
    if (l > R || r < L) {
      return 0;
    } else if (L <= l && r <= R) {
      return tr[i].sum;
    } else {
      push(l, r, i);
      return query(l, (l + r) / 2, i * 2, L, R) +
             query((l + r) / 2 + 1, r, i * 2 + 1, L, R);
    }
  }
} le, ri;
struct fenwick_tree {
  int bit[N];
  void update(int u, int v) {
    for (; u <= n; u += u & -u) {
      bit[u] += v;
    }
  }
  int query(int u) {
    int ret = 0;
    for (; u > 0; u -= u & -u) {
      ret += bit[u];
    }
    return ret;
  }
} chk;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  le.build(1, n, 1, N);
  ri.build(1, n, 1, -N);
  for (int i = 1; i <= n; i++) {
    cin >> u;
    a[u] = i;
  }
  for (int i = 1; i <= n; i++) {
    int pos = a[i];
    le.update(1, n, 1, pos, pos, 1, CHMIN);
    ri.update(1, n, 1, pos, pos, i, CHMAX);
    le.update(1, n, 1, pos + 1, n, 1, ADD);
    ri.update(1, n, 1, pos + 1, n, 1, ADD);
    int cl = chk.query(pos);
    ri.update(1, n, 1, 1, pos - 1, cl, CHMIN);
    le.update(1, n, 1, pos + 1, n, cl + 2, CHMAX);
    chk.update(pos, 1);
    cout << ri.query(1, n, 1, 1, n) - le.query(1, n, 1, 1, n) +
                2LL * N * (n - i) + i
         << '\n';
  }
}
