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 maxn = 150505;
struct tree_array {
  int c[maxn], tot;
  inline void Add(int x, int val) {
    if (!x) return;
    for (int i = x; i <= tot; i += (i & -i)) c[i] += val;
  }
  int getsum(int st, int ed) {
    int ret = 0;
    for (int i = ed; i; i -= (i & -i)) ret += c[i];
    if (st >= 0)
      for (int i = st; i; i -= (i & -i)) ret -= c[i];
    return ret;
  }
  inline void clr() {
    tot = 0;
    memset(c, 0, sizeof(c));
  }
} A, B;
int ch[maxn], df[maxn], n, a[maxn], p[maxn];
struct Node {
  Node *l, *r;
  int st, ed, mx, se, cnt, lazy;
} pool[303030], *root;
int z = 0, tot = 0;
inline void down(Node *now) {
  now->l->mx = min(now->mx, now->l->mx);
  now->r->mx = min(now->mx, now->r->mx);
}
inline void update(Node *now) {
  if (now->l->mx == now->r->mx) {
    now->cnt = now->l->cnt + now->r->cnt;
    now->se = max(now->l->se, now->r->se);
    now->mx = now->l->mx;
  } else if (now->l->mx > now->r->mx) {
    now->mx = now->l->mx;
    now->cnt = now->l->cnt;
    now->se = max(now->r->mx, now->l->se);
  } else {
    now->mx = now->r->mx;
    now->cnt = now->r->cnt;
    now->se = max(now->l->mx, now->r->se);
  }
}
inline void Build(Node *&now, int st, int ed) {
  now = &pool[++z];
  now->st = st;
  now->ed = ed;
  if (st == ed) {
    now->se = -1e9;
    now->mx = 0;
    now->cnt = 1;
    return;
  }
  Build(now->l, st, (st + ed) / 2);
  Build(now->r, (st + ed) / 2 + 1, ed);
  update(now);
}
inline void clr(Node *now) {
  if (now->st == now->ed) {
    now->se = -1e9;
    now->mx = 0;
    now->cnt = 1;
  } else {
    clr(now->l);
    clr(now->r);
    update(now);
  }
}
inline void ins(Node *now, int pos, int val) {
  if (now->st > pos || now->ed < pos) return;
  if (now->st == now->ed) {
    ch[++tot] = now->mx;
    df[tot] = -1;
    now->mx = val;
    ch[++tot] = now->mx;
    df[tot] = 1;
    return;
  }
  down(now);
  ins(now->l, pos, val);
  ins(now->r, pos, val);
  update(now);
}
inline void IntervalMin(Node *now, int st, int ed, int val) {
  if (now->st > ed || now->ed < st || now->mx <= val) return;
  if (now->st >= st && now->ed <= ed && now->se < val) {
    ch[++tot] = now->mx;
    df[tot] = -now->cnt;
    now->mx = val;
    ch[++tot] = now->mx;
    df[tot] = now->cnt;
    return;
  }
  down(now);
  IntervalMin(now->l, st, ed, val);
  IntervalMin(now->r, st, ed, val);
  update(now);
}
set<int> S, T;
long long ans[maxn] = {0};
inline void solve(int X) {
  S.clear();
  T.clear();
  A.clr();
  B.clr();
  A.tot = B.tot = n;
  if (X == 0)
    Build(root, 1, n);
  else
    clr(root);
  S.insert(p[1]);
  T.insert(p[1] - 1);
  A.Add(p[1], 1);
  B.Add(p[1] - 1, -1);
  B.Add(p[1], 1);
  int mx = p[1];
  ins(root, p[1], p[1]);
  long long tmp = 1;
  ans[1] += 1;
  set<int>::iterator it;
  for (int i = 2; i <= n; i++) {
    tot = 0;
    A.Add(p[i], 1);
    tmp += B.getsum(p[i] - 1, n);
    it = S.lower_bound(p[i]);
    if (it == S.end()) {
      ins(root, *S.begin(), p[i]);
      mx = p[i];
    } else if (it == S.begin()) {
      ins(root, p[i], mx);
      B.Add(p[i] - 1, -1);
      T.insert(p[i] - 1);
    } else {
      ins(root, *it, mx);
      if (T.find(*it - 1) == T.end()) {
        B.Add(*it - 1, -1);
        tmp -= A.getsum(0, *it - 1);
        T.insert(*it - 1);
      }
      it--;
      IntervalMin(root, 1, p[i] - 1, *it);
      ins(root, *S.begin(), mx);
    }
    int ttt;
    for (int j = 1; j <= tot; j++) {
      ttt = A.getsum(0, ch[j]);
      tmp += (long long)ttt * df[j];
      B.Add(ch[j], df[j]);
    }
    S.insert(p[i]);
    ans[i] += tmp;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  solve(0);
  for (int i = 1; i <= n; i++) {
    p[i] = n + 1 - p[i];
  }
  solve(1);
  for (int i = 1; i <= n; i++) {
    ans[i] -= i;
    printf("%I64d\n", ans[i]);
  }
  return 0;
}
