time limit per test2 seconds

memory limit per test256 megabytes

There are n walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the n-th walrus stands at the beginning of the queue. The i-th walrus has the age equal to a

The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that ai > aj. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of walruses in the queue. The second line contains integers ai(1 ≤ ai ≤ 109).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.

Output

Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

Examples

Input

6
10 8 5 3 50 45

Output

2 1 0 -1 0 -1 

Input

7
10 4 6 3 2 8 15

Output

4 2 1 0 -1 -1 -1 

Input

5
10 3 1 10 11

Output

1 0 -1 -1 -1 

#include<bits/stdc++.h>
using namespace std;
const int maxn=100010;
vector<int>v1,v2;
int a[maxn];
int ans[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        v1.clear();
        v2.clear();
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=n;i>=1;i--)
        {
            if(v1.size()==0||v1.back()>=a[i])
            {
                v1.push_back(a[i]);
                v2.push_back(i);
                ans[i]=-1;
            }
            else
            {//c.rbegin() 返回一個逆序迭代器，它指向容器c的最後一個元素
             //c.rend() 返回一個逆序迭代器，它指向容器c的第一個元素前面的位置
                int j=lower_bound(v1.rbegin(),v1.rend(),a[i])-v1.rbegin();//我們按照正序排序找a[i]的下限位置。
                j=v1.size()-j;//v1中有j個大於等於a[i]的。
                ans[i]=v2[j]-i-1;//第j+1是v2[j]，此時是最後面的比a[i]小的數。
            }
        }
        printf("%d",ans[1]);
        for(int i=2;i<=n;i++)
            printf(" %d",ans[i]);
        printf("\n");
    }
}

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