codeforces-540C

题目连接：http://codeforces.com/problemset/problem/540/C

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 (r₂, 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 6X...XX...XX..X..X.......1 62 2

Output

YES

Input

5 4.X.....XX.X......XX.5 31 1

Output

NO

Input

4 7..X.XX..XX..X.X...X..X......2 21 6

Output

YES

题目大意：给定一个矩阵，由"."和"X"组成，分别代表完整的冰块和裂缝的冰块，完整的冰块走过一次就会变成裂缝的冰块，裂缝的冰块走过就会掉下去，给定起点和终点，在保证不会掉下去的情况下，问能否从起点走到终点恰好在终点掉下去。解题思路：dfs裸题，因为不需要找多条路径，所以对于每个节点，若经过它的某一条路径能寻找到结果，则必定会在第一次经过该点时就找到，所以不需要重置已经走过的路线，所以暴力dfs即可，每次走可走的完整冰块，完整冰块走过以后置为裂缝冰块，最后得出结果即可。

代码如下：

#include<bits/stdc++.h>

using namespace std;

][];
int n,m,sx,sy,ex,ey;
]= {,,,-,,};
]= {,,,,-,};
;

void dfs(int x,int y)
{
    cout<<"x="<<x<<"y="<<y<<endl;
    if(g[x][y]=='.')
        g[x][y]='X';
    ; u<=; u++)
    {
        int a=x+fx[u];
        int b=y+fy[u];
        &&b<=m&&b>=)
        {
            if(g[a][b]=='.')
            {
                dfs(a,b);
            }
            else
            {
                if(a==ex&&b==ey)
                {
                    flag=;
                    return;
                }
                else
                    continue;
            }
        }
    }
    return;
}

int main()
{
    cin>>n>>m;
    ];
    ; i<=n; i++)
    {
        scanf("%s",in);
        ; j<=m; j++)
            g[i][j]=];
    }
    cin>>sx>>sy;
    cin>>ex>>ey;
    dfs(sx,sy);
    if(flag)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
}