C. Ice Cave time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.

The level of the cave where you are is a rectangular square grid of n rows and m columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.

Let's number the rows with integers from 1 to n from top to bottom and the columns with integers from 1 to m from left to right. Let's denote a cell on the intersection of the r-th row and the c-th column as (r, c).

You are staying in the cell (r₁, c₁) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (r

₂, c₂) since the exit to the next level is there. Can you do this?

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 500) — the number of rows and columns in the cave description.

Each of the next n lines describes the initial state of the level of the cave, each line consists of m characters ".

" (that is, intact ice) and "X" (cracked ice).

The next line contains two integers, r₁ and c₁ (1 ≤ r₁ ≤ n, 1 ≤ c₁ ≤ m) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (r₁, c₁), that is, the ice on the starting cell is initially cracked.

The next line contains two integers r₂ and c₂ (1 ≤ r₂ ≤ n, 1 ≤ c₂ ≤ m) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.

Output

If you can reach the destination, print 'YES', otherwise print 'NO'.

Examples Input

4 6
X...XX
...XX.
.X..X.
......
1 6
2 2

Output

YES

Input

5 4
.X..
...X
X.X.
....
.XX.
5 3
1 1

Output

NO

Input

4 7
..X.XX.
.XX..X.
X...X..
X......
2 2
1 6

Output

YES

Note

In the first sample test one possible path is:

After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended.

題目大意：

從起點走到終點，走到點的話，點會變成X，如果走到了X的話，就會掉下去。如果掉下去的點是終點，那麼輸出YES,否則輸出NO。

思路：因為點的狀態會有變化，所以最開始以為廣搜要有伴隨結構體走的圖，沒敢敲BFS,敲的DFS，果斷TLE。。。。。。其實仔細想想完全不需要伴隨結構體走的map，如果有走到點的地方讓他能變成X，那麼無非必要討論其他到這裡會不會有點變叉的問題，還是欠缺了思考。

AC程式碼：

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
struct zuobiao
{
    int x,y;
}now,nex;
int sx,sy,ex,ey;
char a[505][505];
int fx[4]={0,0,-1,1};
int fy[4]={1,-1,0,0};
int n,m;
int ok;
void bfs(int x,int y)
{
    now.x=x;
    now.y=y;
    queue<zuobiao >s;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();
        s.pop();
        for(int i=0;i<4;i++)
        {
            nex.x=fx[i]+now.x;
            nex.y=fy[i]+now.y;
            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m)
            {
                if(a[nex.x][nex.y]=='X')
                {
                    if(nex.x==ex&&nex.y==ey)
                    {
                        printf("YES\n");
                        return ;
                    }
                }
                else
                {
                    a[nex.x][nex.y]='X';
                    s.push(nex);
                }
            }
        }
    }
    printf("NO\n");
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
        }
        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
        sx--;sy--;ex--;ey--;
        ok=0;
        bfs(sx,sy);
    }
}