Codeforces Round #456 (Div. 2) D. Fishes

While Grisha was celebrating New Year with Ded Moroz, Misha gifted Sasha a small rectangular pond of sizen × m, divided into cells of size1 × 1, inhabited by tiny evil fishes (no more than one fish per cell, otherwise they'll strife!).

The gift bundle also includes a square scoop of size r × r, designed for fishing. If the lower-left corner of the scoop-net is located at cell(x, y), all fishes inside the square(x, y)...(x + r - 1, y + r - 1) get caught. Note that the scoop-net should lie completely inside the pond when used.

Unfortunately, Sasha is not that skilled in fishing and hence throws the scoop randomly. In order to not frustrate Sasha, Misha decided to releasek fishes into the empty pond in such a way that the expected value of the number of caught fishes is as high as possible. Help Misha! In other words, putk fishes in the pond into distinct cells in such a way that when the scoop-net is placed into a random position among(n - r + 1)·(m - r + 1) possible positions, the average number of caught fishes is as high as possible.

Input

The only line contains four integers n, m, r, k (1 ≤ n, m ≤ 10⁵,1 ≤ r ≤ min(n, m),1 ≤ k ≤ min(n·m, 10⁵)).

Output

Print a single number — the maximum possible expected number of caught fishes.

You answer is considered correct, is its absolute or relative error does not exceed10^- 9. Namely, let your answer bea, and the jury's answer be b. Your answer is considered correct, if .

Examples

Input

3 3 2 3

Output

2.0000000000

Input

12 17 9 40

Output

32.8333333333

Note

In the first example you can put the fishes in cells (2, 1),(2, 2), (2, 3). In this case, for any of four possible positions of the scoop-net (highlighted with light green), the number of fishes inside is equal to two, and so is the expected value.

题目大意：用r*r的矩阵去套k条鱼，问期望值是多少？

思路：中间放一条鱼对答案贡献肯定最大，根据对称性，上下左右可得下一个最大值，向四个方向BFS，贡献值逐级递减

#include <bits/stdc++.h>
typedef long long ll;

using namespace std;

int n,m,r,k;
ll sum;
map<pair<int,int>,bool> vis;
int dir[4][2]={0,1, 0,-1, 1,0, -1,0};

struct node{
    int x,y;
    ll v;
    bool operator <(const node & obj)    const{ //为了排序
       return v<obj.v;
    }
};

ll cul(int x,int y){
    int xmin,xmax,ymin,ymax;
    xmin=max(x,r),xmax=min(x+r-1,n);
    ymin=max(y,r),ymax=min(y+r-1,m);
    return 1LL*(xmax-xmin+1)*(ymax-ymin+1);
}

ll solve(){
    node temp;
    temp.x=(n+1)/2,temp.y=(m+1)/2;
    temp.v=cul(temp.x,temp.y);
    vis[make_pair(temp.x,temp.y)]=true;
    priority_queue<node> que;
    que.push(temp);
    while(!que.empty()){
        node t,tt;
        t=que.top();
        que.pop();
        sum+=t.v;
        k--;
        if(k==0)    break;
        for(int i=0;i<4;i++){
            int nx=dir[i][0]+t.x;
            int ny=dir[i][1]+t.y;
            if(!vis[make_pair(nx,ny)]&&nx>=1&&nx<=n&&ny>=1&&ny<=m){
                tt.x=nx,tt.y=ny,tt.v=cul(nx,ny);
                que.push(tt);
                vis[make_pair(nx,ny)]=true;

            }

        }
        //puts("");
    }
    return sum;
}

int main(void){
    cin>>n>>m>>r>>k;
    ll tot=1LL*(n-r+1)*(m-r+1);
    ll sum=solve();
    printf("%.12f\n",sum*1.0/tot);
}