Funny Feature

In the process of preparation to Halloween it was decided to plant some pumpkins on the n × m meters rectangular platform. The platform is divided to n × midentical square cells with 1 meter length sides.

The arrangement of pumpkins was carefully prepared, and you are given the resulting plan. For each cell of this platform it is given how many pumpkins should be planted in it. All these quantities appear to be from 1 to 5, inclusive. 

Special pumpkin-landing machine was bought to plant Halloween symbols. It can perform a simple operation: plant pumpkin to the cell, specified by its coordinates (the first coordinate ranges 1 to n, and the second one ranges 1 to m). 

At the last moment an unpleasant feature of the machine was discovered. Every time this machine plants a pumpkin to the specified cell, one more pumpkin is also planted to all cells, which are adjacent to the specified one and already have at least one pumpkin in it. One cell is called adjacent to another one if they share a side.

Besides for technical reasons you cannot specify the same cell twice.

Now Halloween celebration is under the threat of failure. You are asked to write the program which finds the sequence of n × m operations leading to demanded landing of pumpkins, or informs that it is impossible.

Input

The first line of input contains two integers n and m  — sizes of the field (1 ≤ n,m ≤ 200). Next n lines contain m integers from 1 to 5  — how many pumpkins should be planted in each cell of the platform. Numbers in lines are separated by single spaces.

Output

If the solution exists, print n × m lines with two numbers in each: a line and a column numbers, where the next pumpkin should be landed. If there are multiple solutions, print any of them.

If the solution does not exist, print a single line "No solution" (quotes for clarity).

Example(s)

sample input	sample output
1 2 1 2	1 2 1 1

sample input	sample output
1 3 1 3 1	1 2 1 3 1 1

sample input	sample output
3 3 2 4 2 2 1 2 3 2 3	1 2 3 1 3 3 1 1 1 3 3 2 2 1 2 3 2 2

sample input	sample output
2 1 1 1	No solution

#include<queue>
#include<stack>
#include<iostream>
#include<cstdlib>
using namespace std;
const int maxn=200+10;
const int di[]={-1,0,1,0},dj[]={0,1,0,-1};
int map[maxn][maxn],n,m;
bool mark[maxn][maxn]={0};
struct node
{
int x,y;
node(int x,int y):x(x),y(y){}
void show()
{
cout<<x<<" "<<y<<endl;
}
};
void Cant()
{
cout<<"No solution"<<endl;
exit(0);
}
int main()
{
cin>>n>>m;    
queue<node> Q;
stack<node> S;
for(int i = 0; i <= n+1;i++) for(int j = 0; j<=m+1;j++) map[i][j]=10;
for(int i = 1; i<=n;i++)
REP(int j =1; j<=m;j++)
{
cin>>map[i][j];    
if(map[i][j]==1)
Q.push(node(i,j));
}
while(Q.size())
{
node cur=Q.front();Q.pop();
S.push(cur);
mark[cur.x][cur.y]=true;        
for(int k =0; k <=3;k++)
{
int t=di[k]+cur.x,u=dj[k]+cur.y;
if(!mark[t][u]&&map[t][u]!=10)
{
map[t][u]–;
if(!map[t][u])
Cant();
if(map[t][u]==1)
Q.push(node(t,u));
}
}
}
if(S.size()!=n*m)
Cant();
while(S.size())
{
S.top().show();S.pop();
}
}