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;
inline int read_int() {
  int t = 0;
  bool sign = false;
  char c = getchar();
  while (!isdigit(c)) {
    sign |= c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    t = (t << 1) + (t << 3) + (c & 15);
    c = getchar();
  }
  return sign ? -t : t;
}
inline long long read_LL() {
  long long t = 0;
  bool sign = false;
  char c = getchar();
  while (!isdigit(c)) {
    sign |= c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    t = (t << 1) + (t << 3) + (c & 15);
    c = getchar();
  }
  return sign ? -t : t;
}
inline char get_char() {
  char c = getchar();
  while (c == ' ' || c == '\n' || c == '\r') c = getchar();
  return c;
}
inline void write(long long x) {
  register char c[21], len = 0;
  if (!x) return putchar('0'), void();
  if (x < 0) x = -x, putchar('-');
  while (x) c[++len] = x % 10, x /= 10;
  while (len) putchar(c[len--] + 48);
}
inline void space(long long x) { write(x), putchar(' '); }
inline void enter(long long x) { write(x), putchar('\n'); }
const int MAXN = 1.5e5 + 5, Inf = 2e9;
int a[MAXN];
int lef[MAXN << 2], rig[MAXN << 2], maxv[MAXN << 2], maxc[MAXN << 2],
    secv[MAXN << 2], sz[MAXN << 2];
int lazy1[MAXN << 2], lazy2[MAXN << 2];
long long sum[MAXN << 2];
void push_up(int k) {
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
  sz[k] = sz[k << 1] + sz[k << 1 | 1];
  if (maxv[k << 1] == maxv[k << 1 | 1]) {
    maxv[k] = maxv[k << 1];
    maxc[k] = maxc[k << 1] + maxc[k << 1 | 1];
    secv[k] = max(secv[k << 1], secv[k << 1 | 1]);
  } else if (maxv[k << 1] > maxv[k << 1 | 1]) {
    maxv[k] = maxv[k << 1];
    maxc[k] = maxc[k << 1];
    secv[k] = max(secv[k << 1], maxv[k << 1 | 1]);
  } else {
    maxv[k] = maxv[k << 1 | 1];
    maxc[k] = maxc[k << 1 | 1];
    secv[k] = max(maxv[k << 1], secv[k << 1 | 1]);
  }
}
void push_add(int k, int v) {
  sum[k] += 1LL * sz[k] * v;
  maxv[k] += v, lazy1[k] += v;
  if (secv[k] != -Inf) secv[k] += v;
  if (lazy2[k] != -Inf) lazy2[k] += v;
}
void push_min(int k, int v) {
  if (v >= maxv[k]) return;
  sum[k] -= 1LL * maxc[k] * (maxv[k] - v);
  maxv[k] = lazy2[k] = v;
}
void push_down(int k) {
  if (lazy1[k]) {
    push_add(k << 1, lazy1[k]);
    push_add(k << 1 | 1, lazy1[k]);
    lazy1[k] = 0;
  }
  if (lazy2[k] != -Inf) {
    push_min(k << 1, lazy2[k]);
    push_min(k << 1 | 1, lazy2[k]);
    lazy2[k] = -Inf;
  }
}
void build(int k, int L, int R) {
  lef[k] = L, rig[k] = R, lazy1[k] = 0, lazy2[k] = -Inf;
  if (L == R) {
    sum[k] = maxv[k] = sz[k] = maxc[k] = 0;
    secv[k] = -Inf;
    return;
  }
  int M = L + R >> 1;
  build(k << 1, L, M);
  build(k << 1 | 1, M + 1, R);
  push_up(k);
}
void update_add(int k, int L, int R, int v) {
  if (L <= lef[k] && rig[k] <= R) return push_add(k, v);
  push_down(k);
  int mid = lef[k] + rig[k] >> 1;
  if (mid >= L) update_add(k << 1, L, R, v);
  if (mid < R) update_add(k << 1 | 1, L, R, v);
  push_up(k);
}
void update_min(int k, int L, int R, int v) {
  if (maxv[k] <= v) return;
  if (L <= lef[k] && rig[k] <= R && secv[k] < v) return push_min(k, v);
  push_down(k);
  int mid = lef[k] + rig[k] >> 1;
  if (mid >= L) update_min(k << 1, L, R, v);
  if (mid < R) update_min(k << 1 | 1, L, R, v);
  push_up(k);
}
void insert_node(int k, int pos, int v) {
  if (lef[k] == rig[k]) {
    sum[k] = maxv[k] = v;
    maxc[k] = sz[k] = 1;
    return;
  }
  push_down(k);
  int mid = lef[k] + rig[k] >> 1;
  if (mid >= pos)
    insert_node(k << 1, pos, v);
  else
    insert_node(k << 1 | 1, pos, v);
  push_up(k);
}
int query_sz(int k, int L, int R) {
  if (L <= lef[k] && rig[k] <= R) return sz[k];
  push_down(k);
  int mid = lef[k] + rig[k] >> 1;
  int s = 0;
  if (mid >= L) s += query_sz(k << 1, L, R);
  if (mid < R) s += query_sz(k << 1 | 1, L, R);
  return s;
}
int p[MAXN];
long long ans[MAXN];
int main() {
  int n = read_int();
  for (int i = (1); i <= (n); ++i) p[read_int()] = i;
  build(1, 1, n);
  for (int i = (1); i <= (n); ++i) {
    insert_node(1, p[i], i + 1);
    if (p[i] < n) update_add(1, p[i] + 1, n, 1);
    if (p[i] > 1) update_min(1, 1, p[i] - 1, query_sz(1, 1, p[i]));
    ans[i] += sum[1];
  }
  build(1, 1, n);
  for (int i = (1); i <= (n); ++i) p[i] = n + 1 - p[i];
  for (int i = (1); i <= (n); ++i) {
    insert_node(1, p[i], i + 1);
    if (p[i] < n) update_add(1, p[i] + 1, n, 1);
    if (p[i] > 1) update_min(1, 1, p[i] - 1, query_sz(1, 1, p[i]));
    ans[i] += sum[1];
  }
  for (int i = (1); i <= (n); ++i) enter(ans[i] - 1LL * i * (i + 2));
  return 0;
}
