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 MN = 1.5e5 + 100;
using ll = long long;
int N, NN, a[MN], b[MN];
ll fans[MN];
ll ans;
struct BIT {
 public:
  int a[MN];
  void reset() { memset(a, 0, sizeof(a)); }
  void upd(int x, int v) {
    assert(x >= 0);
    for (++x; x <= N + 10; x += -x & x) a[x - 1] += v;
  }
  int qry(int x) {
    assert(x >= 0);
    int r = 0;
    for (++x; x; x -= -x & x) r += a[x - 1];
    return r;
  }
};
BIT vals, cnts;
struct segtree {
 public:
  int m[1 << 19], m2[1 << 19], c[1 << 19], z[1 << 19];
  void reset() {
    memset(m2, -1, sizeof(m2));
    memset(z, -1, sizeof(z));
    memset(c, 0, sizeof(c));
    memset(m, -1, sizeof(m));
  }
  void down(int n) {
    if (!~z[n]) return;
    for (int x : {n << 1, n << 1 | 1})
      if (m[x] > z[n]) z[x] = m[x] = z[n];
    z[n] = -1;
  }
  void up(int n) {
    if (m[n << 1] > m[n << 1 | 1])
      m[n] = m[n << 1], c[n] = c[n << 1],
      m2[n] = std::max(m2[n << 1], m[n << 1 | 1]);
    else if (m[n << 1] < m[n << 1 | 1])
      m[n] = m[n << 1 | 1], c[n] = c[n << 1 | 1],
      m2[n] = std::max(m2[n << 1 | 1], m[n << 1]);
    else
      m[n] = m[n << 1], c[n] = c[n << 1] + c[n << 1 | 1],
      m2[n] = std::max(m2[n << 1], m2[n << 1 | 1]);
  }
  void set(int n, int l, int r, int x, int v) {
    if (r - l > 1) {
      int m = l + (r - l) / 2;
      down(n);
      if (x < m)
        set(n << 1, l, m, x, v);
      else
        set(n << 1 | 1, m, r, x, v);
      up(n);
    } else {
      if (c[n]) cnts.upd(m[n], -1), ans -= vals.qry(m[n]);
      m[n] = v, c[n] = 1;
      cnts.upd(m[n], 1);
      ans += vals.qry(m[n]);
    }
  }
  void upd(int n, int l, int r, int ql, int qr, int v) {
    if (v >= m[n]) return;
    if (v > m2[n] && ql <= l && r <= qr) {
      cnts.upd(m[n], -c[n]);
      ans -= static_cast<ll>(c[n]) * vals.qry(m[n]);
      m[n] = v;
      z[n] = v;
      cnts.upd(m[n], c[n]);
      ans += static_cast<ll>(c[n]) * vals.qry(m[n]);
      return;
    }
    assert(r - l > 1);
    int m = l + (r - l) / 2;
    down(n);
    if (ql < m) upd(n << 1, l, m, ql, qr, v);
    if (m < qr) upd(n << 1 | 1, m, r, ql, qr, v);
    up(n);
  }
};
segtree st;
std::set<int> use;
bool start[MN];
void solve() {
  st.reset();
  vals.reset();
  cnts.reset();
  use.clear();
  memset(start, 0, sizeof(start));
  ans = 0LL;
  for (int i = 0, x; i < N; ++i) {
    st.upd(1, 0, NN, 0, b[i], b[i] - 1);
    std::set<int>::iterator it = use.upper_bound(b[i]);
    if (it != use.end()) {
      st.set(1, 0, NN, *it, N);
      if (!start[*it])
        cnts.upd(*it - 1, -1), start[*it] = 1, ans -= vals.qry(*it - 1);
    }
    use.insert(b[i]);
    st.set(1, 0, NN, *use.begin(), N);
    if (!start[*use.begin()])
      cnts.upd(*use.begin() - 1, -1), start[*use.begin()] = 1,
                                      ans -= vals.qry(*use.begin() - 1);
    vals.upd(b[i], 1);
    ans += cnts.qry(N) - cnts.qry(b[i] - 1);
    fans[i] += ans;
  }
}
int main(void) {
  scanf("%d", &N);
  NN = 1 << 18;
  for (int i = 0; i < N; ++i) scanf("%d", a + i), b[a[i] - 1] = i + 1;
  solve();
  for (int i = 0; i < N; ++i) b[i] = N + 1 - b[i];
  solve();
  for (int i = 0; i < N; ++i) printf("%lld\n", fans[i] - i - 1);
  return 0;
}
