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;
namespace Fread {
const int MAXN = 1 << 20;
char buf[MAXN], *S, *T;
inline char getchar() {
  if (S == T) {
    T = (S = buf) + fread(buf, 1, MAXN, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
}  // namespace Fread
inline int read() {
  int f = 1, x = 0;
  char ch = Fread::getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = Fread::getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = Fread::getchar();
  }
  return x * f;
}
inline long long readll() {
  long long f = 1, x = 0;
  char ch = Fread::getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = Fread::getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = Fread::getchar();
  }
  return x * f;
}
const int MAXN = 150000, INF = 1e9;
int n, a[MAXN + 5], p[MAXN + 5];
long long ans[MAXN + 5];
vector<pair<int, long long> > ch;
int mx[MAXN * 4 + 5], se[MAXN * 4 + 5], mxc[MAXN * 4 + 5];
inline void push_up(int p) {
  int ls = p << 1, rs = p << 1 | 1;
  if (mx[ls] < mx[rs]) {
    mx[p] = mx[rs];
    se[p] = max(mx[ls], se[rs]);
    mxc[p] = mxc[rs];
  } else if (mx[ls] > mx[rs]) {
    mx[p] = mx[ls];
    se[p] = max(se[ls], mx[rs]);
    mxc[p] = mxc[ls];
  } else {
    mx[p] = mx[ls];
    se[p] = max(se[ls], se[rs]);
    mxc[p] = mxc[ls] + mxc[rs];
  }
}
void build(int p, int l, int r) {
  if (l == r) {
    mx[p] = 0;
    se[p] = -INF;
    mxc[p] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push_up(p);
}
void push_down(int p) {
  mx[p << 1] = min(mx[p << 1], mx[p]);
  mx[p << 1 | 1] = min(mx[p << 1 | 1], mx[p]);
}
void modify1(int p, int l, int r, int ql, int qr, int x) {
  if (mx[p] <= x) return;
  if (ql <= l && qr >= r && x > se[p]) {
    if (mx[p]) ch.push_back(make_pair(mx[p], -mxc[p]));
    mx[p] = x;
    ch.push_back(make_pair(mx[p], mxc[p]));
    return;
  }
  push_down(p);
  int mid = (l + r) >> 1;
  if (ql <= mid) modify1(p << 1, l, mid, ql, qr, x);
  if (qr > mid) modify1(p << 1 | 1, mid + 1, r, ql, qr, x);
  push_up(p);
}
void modify2(int p, int l, int r, int pos, int x) {
  if (l == r) {
    if (mx[p]) ch.push_back(make_pair(mx[p], -mxc[p]));
    mx[p] = x;
    ch.push_back(make_pair(mx[p], mxc[p]));
    return;
  }
  push_down(p);
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify2(p << 1, l, mid, pos, x);
  else
    modify2(p << 1 | 1, mid + 1, r, pos, x);
  push_up(p);
}
struct Fenwick {
  long long c[MAXN + 5];
  void modify(int p, long long x) {
    if (!p) return;
    for (; p <= n; p += (p & (-p))) c[p] += x;
  }
  long long query(int p) {
    long long res = 0;
    for (; p; p -= (p & (-p))) res += c[p];
    return res;
  }
  long long query(int l, int r) {
    if (l > r) return 0;
    return query(r) - query(l - 1);
  }
  void init() { memset(c, 0, sizeof(c)); }
  Fenwick() {}
} T1, T2;
void solve() {
  build(1, 1, n);
  T1.init();
  T2.init();
  set<int> s, t;
  s.insert(p[1]);
  t.insert(p[1]);
  modify2(1, 1, n, p[1], p[1]);
  T1.modify(p[1], 1);
  T2.modify(p[1] - 1, -1);
  long long tot = 0;
  int mx = p[1];
  for (int i = 2; i <= n; ++i) {
    set<int>::iterator it = s.lower_bound(p[i]);
    int nxt = 0;
    if (it != s.end())
      nxt = (*it);
    else
      mx = p[i];
    if (it == s.begin()) {
      tot--;
      t.insert(p[i]);
      T2.modify(p[i] - 1, -1);
      modify2(1, 1, n, p[i], mx);
    } else {
      --it;
      int pre = (*it);
      modify1(1, 1, n, 1, pre, pre);
      modify2(1, 1, n, *s.begin(), mx);
      if (nxt) {
        modify2(1, 1, n, nxt, mx);
        if (t.find(nxt) == t.end()) {
          tot -= T1.query(nxt);
          t.insert(nxt);
          T2.modify(nxt - 1, -1);
        }
      }
    }
    s.insert(p[i]);
    for (int j = 0; j < (int)ch.size(); ++j) {
      tot += T1.query(ch[j].first) * ch[j].second;
      T2.modify(ch[j].first, ch[j].second);
    }
    T1.modify(p[i], 1);
    tot += T2.query(p[i], n);
    ch.clear();
    ans[i] += tot;
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read(), p[a[i]] = i;
  solve();
  for (int i = 1; i <= n; ++i) p[i] = n - p[i] + 1;
  solve();
  for (int i = 2; i <= n; ++i) ans[i]--;
  ans[1] = 1;
  for (int i = 1; i <= n; ++i) printf("%lld\n", ans[i]);
  return 0;
}
