ICE CAVE（BFS搜索（模拟））

Description

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'.

Sample Input

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


如题，每走一步，走过的那个地方变成破碎的冰块，如果你再走就会掉下去，你不能在原地跳动使冰块改变状态，而你最终的目地不只是找到目的地，而且要让它为X然后再次进入

原先我还想着要分情况考虑来着，比如目的地原来是X还是。，然后判断条件就变了，之后又要计数，标不标记又是个问题，又考虑从原点到原点。。。。。。。。。。。

最后一塌糊涂，要考虑的太多，问题就很复杂，其实大可不必，只要将到达那一点的条件加一句地图上该点是否为X，其余的模拟实现，走到一点，那一点就变成X。。。。。。

#include"iostream" #include"algorithm" #include"cstring" #include"cstdio" #include"queue" using namespace std; const int maxn=510; char a[maxn][maxn]; int m,n, book[maxn][maxn]; int nex[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; int sp,sq,p,q,tp,tq; struct node { int x; int y; }; int BFS() { queue<struct node>que; struct node c,e,t; c.x=sp; c.y=sq; que.push(c); while(!que.empty()) { e=que.front(); que.pop(); for(int k=0;k<=3;k++) { t.x=e.x+nex[k][0]; t.y=e.y+nex[k][1]; if(t.x<0||t.x>=m||t.y<0||t.y>=n) continue; if(t.x==p&&t.y==q&&a[t.x][t.y]=='X') { return 1; } // continue; // book[t.x][t.y]=1; if(a[t.x][t.y]=='X') continue; a[t.x][t.y]='X'; que.push(t); } } return 0; } int main() { while(cin>>m>>n&&m) { for(int i=0;i<m;i++) { scanf("%s",a[i]); } memset(book,0,sizeof(book)); cin>>sp>>sq; cin>>p>>q; sp=sp-1; sq=sq-1; p=p-1; q=q-1; if(BFS()) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }