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 = 1.5e5 + 5;
int a[maxn];
struct JlsTree {
  int mx1[maxn << 2], mx2[maxn << 2], ge[maxn << 2], gee[maxn << 2],
      ad[maxn << 2], mn[maxn << 2];
  long long sum[maxn << 2];
  void pushup(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    if (mx1[rt << 1] == mx1[rt << 1 | 1]) {
      mx1[rt] = mx1[rt << 1];
      ge[rt] = ge[rt << 1] + ge[rt << 1 | 1];
      mx2[rt] = max(mx2[rt << 1], mx2[rt << 1 | 1]);
    } else if (mx1[rt << 1] < mx1[rt << 1 | 1]) {
      mx1[rt] = mx1[rt << 1 | 1];
      ge[rt] = ge[rt << 1 | 1];
      mx2[rt] = max(mx1[rt << 1], mx2[rt << 1 | 1]);
    } else {
      mx1[rt] = mx1[rt << 1];
      ge[rt] = ge[rt << 1];
      mx2[rt] = max(mx1[rt << 1 | 1], mx2[rt << 1]);
    }
  }
  void tomin(int rt, int minn) {
    if (mx1[rt] < minn) return;
    if (mx2[rt] < minn) {
      sum[rt] -= ge[rt] * (mx1[rt] - minn);
      mx1[rt] = minn;
      mn[rt] = minn;
    } else {
      if (ad[rt]) {
        mx1[rt << 1] += ad[rt];
        mx2[rt << 1] += ad[rt];
        ad[rt << 1] += ad[rt];
        mn[rt << 1] += ad[rt];
        sum[rt << 1] += 1ll * gee[rt << 1] * ad[rt];
        mx1[rt << 1 | 1] += ad[rt];
        mx2[rt << 1 | 1] += ad[rt];
        ad[rt << 1 | 1] += ad[rt];
        mn[rt << 1 | 1] += ad[rt];
        sum[rt << 1 | 1] += 1ll * gee[rt << 1 | 1] * ad[rt];
        ad[rt] = 0;
      }
      tomin(rt << 1, minn);
      tomin(rt << 1 | 1, minn);
      pushup(rt);
    }
  }
  void pushdown(int rt, int l, int r) {
    int m = l + r >> 1;
    if (ad[rt]) {
      mx1[rt << 1] += ad[rt];
      mx2[rt << 1] += ad[rt];
      ad[rt << 1] += ad[rt];
      mn[rt << 1] += ad[rt];
      sum[rt << 1] += 1ll * gee[rt << 1] * ad[rt];
      mx1[rt << 1 | 1] += ad[rt];
      mx2[rt << 1 | 1] += ad[rt];
      ad[rt << 1 | 1] += ad[rt];
      mn[rt << 1 | 1] += ad[rt];
      sum[rt << 1 | 1] += 1ll * gee[rt << 1 | 1] * ad[rt];
      ad[rt] = 0;
    }
    if (mn[rt] != 0x3f3f3f3f) {
      tomin(rt << 1, mn[rt]);
      tomin(rt << 1 | 1, mn[rt]);
      mn[rt] = 0x3f3f3f3f;
    }
  }
  void build(int rt, int l, int r) {
    mn[rt] = 0x3f3f3f3f;
    if (l == r) return;
    int m = l + r >> 1;
    build(rt << 1, l, m);
    build(rt << 1 | 1, m + 1, r);
  }
  void add(int rt, int l, int r, int L, int R, int x) {
    if (L <= l && r <= R) {
      mx1[rt] += x;
      mx2[rt] += x;
      ad[rt] += x;
      if (mn[rt] != 0x3f3f3f3f) mn[rt] += x;
      sum[rt] += 1ll * x * gee[rt];
      return;
    }
    pushdown(rt, l, r);
    int m = l + r >> 1;
    if (L <= m) add(rt << 1, l, m, L, R, x);
    if (R > m) add(rt << 1 | 1, m + 1, r, L, R, x);
    pushup(rt);
  }
  void givemin(int rt, int l, int r, int L, int R, int x) {
    if (L <= l && r <= R) {
      tomin(rt, x);
      return;
    }
    pushdown(rt, l, r);
    int m = l + r >> 1;
    if (L <= m) givemin(rt << 1, l, m, L, R, x);
    if (R > m) givemin(rt << 1 | 1, m + 1, r, L, R, x);
    pushup(rt);
  }
  void nw(int rt, int l, int r, int x, int num) {
    gee[rt]++;
    if (l == r) {
      sum[rt] = num;
      mx1[rt] = num;
      ge[rt] = 1;
      return;
    }
    pushdown(rt, l, r);
    int m = l + r >> 1;
    if (x <= m)
      nw(rt << 1, l, m, x, num);
    else
      nw(rt << 1 | 1, m + 1, r, x, num);
    pushup(rt);
  }
} jt1, jt2;
int sum[maxn];
int n;
void add(int x) {
  for (int i = x; i <= n; i += (i & (-i))) sum[i]++;
}
int query(int x) {
  int ans = 0;
  for (int i = x; i; i -= (i & (-i))) ans += sum[i];
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[x] = i;
  }
  jt1.build(1, 1, n);
  jt2.build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int y = query(a[i]);
    jt1.givemin(1, 1, n, 1, a[i], y + 1);
    jt1.add(1, 1, n, a[i], n, 1);
    jt1.nw(1, 1, n, a[i], i + 1);
    jt2.givemin(1, 1, n, 1, n - a[i] + 1, i - y);
    jt2.add(1, 1, n, n - a[i] + 1, n, 1);
    jt2.nw(1, 1, n, n - a[i] + 1, i + 1);
    add(a[i]);
    printf("%lld\n", jt1.sum[1] + jt2.sum[1] - 1ll * i * (i + 2));
  }
}
