2201: Chip Play

Time Limit: 2 Sec  Memory Limit: 256 MB Submit: 7  Solved: 6 [Submit][Status][Web Board]

Description

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's consider the following game. We have a rectangular field n

×m in size. Some squares of the field contain chips.

Each chip has an arrow painted on it. Thus, each chip on the field points in one of the following directions: up, down, left or right.

The player may choose a chip and make a move with it.

The move is the following sequence of actions. The chosen chip is marked as the current one. After that the player checks whether there are more chips in the same row (or in the same column) with the current one that are pointed by the arrow on the current chip. If there is at least one chip then the closest of them is marked as the new current chip and the former current chip is removed from the field. After that the check is repeated. This process can be repeated several times. If a new chip is not found, then the current chip is removed from the field and the player's move ends.

By the end of a move the player receives several points equal to the number of the deleted chips.

By the given initial chip arrangement determine the maximum number of points that a player can receive during one move. Also determine the number of such moves.

Input

The first line contains two integers n

and m (1≤n,m,n×m≤5000). Then follow n lines containing m characters each − that is the game field description. "." means that this square is empty. "L", "R", "U", "D" mean that this square contains a chip and an arrow on it says left, right, up or down correspondingly.

It is guaranteed that a field has at least one chip.

Output

Print two numbers − the maximal number of points a player can get after a move and the number of moves that allow receiving this maximum number of points.

Examples

Input

4 4
DRLD
U.UL
.UUR
RDDL

Output

10 1

Input

3 5
.D...
RRRLL
.U...

Output

6 2

Note

In the first sample the maximum number of points is earned by the chip in the position (3,3). You can see its progress at the following picture:

All other chips earn fewer points.

#include<bits/stdc++.h>
using namespace std;
const int maxn=5005;
char mp[maxn][maxn];
int U[maxn],D[maxn],L[maxn],R[maxn];
int n,m;
void init()//初始化 
{
	/*左右*/
	for(int i=0;i<n;i++)
	{
		int l=-1;//l是上一個節點 
		for(int j=0;j<m;j++)
		{
			if(mp[i][j]!='.')
			{
				int now=i*m+j;
				L[now]=l;//當前節點的左節點就是上一個節點 
				if(l!=-1)R[l]=now;//上一個節點的右節點即為當前節點 
				l=now;
			}
		}
		R[l]=-1;
	}
	/*上下-------注意i、j、m、n的變化*/
	for(int i=0;i<m;i++)//相當於把這個棋盤轉了一下 
	{
		int l=-1;
		for(int j=0;j<n;j++)
		{
			if(mp[j][i]!='.')
			{
				int now=j*m+i;
				U[now]=l;
				if(l!=-1)D[l]=now;
				l=now;
			}
		}
		D[l]=-1;
	}
}
int dfs(int x,int y)
{
	int now=x*m+y;
	//刪除當前節點；題意是走過之後就刪掉 
	if(U[now]!=-1)D[U[now]]=D[now]; //如果a結點上面還有節點b，使b跳過a指向a下面這個節點，下面同理，四個方向上實現刪除該節點的功能 
	if(D[now]!=-1)U[D[now]]=U[now];
	if(L[now]!=-1)R[L[now]]=R[now];
	if(R[now]!=-1)L[R[now]]=L[now];
	
	int go;//下一次的位置 
	if(mp[x][y]=='U')go=U[now];
	else if(mp[x][y]=='D')go=D[now];
	else if(mp[x][y]=='L')go=L[now];
	else if(mp[x][y]=='R')go=R[now];
	
	if(go==-1)return 1;//到邊緣，結束，正常退出 
	return dfs(go/m,go%m)+1;//進行下一次搜尋，數量+1； 
}
int main()
{
	int ans=0,num=0;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
		scanf("%s",mp[i]);
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
		{
			if(mp[i][j]!='.')
			{
				init();//初始化十字連結串列（雖然我還不清楚十字連結串列是什麼東東——） 
				int x=dfs(i,j);//cout<<"x="<<x<<endl;
				if(x>ans)ans=x,num=1;//如果最大值更新，數量就歸1； 
				else if(x==ans)num++; 
			}
		}
	printf("%d %d\n",ans,num);
	return 0;
}