Problem: AtCoDeer the deer wants a directed graph that satisfies the following conditions:

* The number of vertices, N, is at most 300.
* There must not be self-loops or multiple edges.
* The vertices are numbered from 1 through N.
* Each edge has either an integer weight between 0 and 100 (inclusive), or a label `X` or `Y`.
* For every pair of two integers (x,y) such that 1 ≤ x ≤ A, 1 ≤ y ≤ B, the shortest distance from Vertex S to Vertex T in the graph where the edges labeled `X` have the weight x and the edges labeled `Y` have the weight y, is d_{x,y}.



Construct such a graph (and a pair of S and T) for him, or report that it does not exist. Refer to Output section for output format.

Constraints

* 1 ≤ A,B ≤ 10
* 1 ≤ d_{x,y} ≤ 100 (1 ≤ x ≤ A, 1 ≤ y ≤ B)
* All input values are integers.

Input

Input is given from Standard Input in the following format:


A B
d_{1,1} d_{1,2} .. d_{1,B}
d_{2,1} d_{2,2} .. d_{2,B}
:
d_{A,1} d_{A,2} .. d_{A,B}


Output

If no graph satisfies the condition, print `Impossible`.

If there exists a graph that satisfies the condition, print `Possible` in the first line. Then, in the subsequent lines, print the constructed graph in the following format:


N M
u_1 v_1 c_1
u_2 v_2 c_2
:
u_M v_M c_M
S T


Here, M is the number of the edges, and u_i, v_i, c_i represent edges as follows: there is an edge from Vertex u_i to Vertex v_i whose weight or label is c_i.

Also refer to Sample Outputs.

Examples

Input

2 3
1 2 2
1 2 3


Output

Possible
3 4
1 2 X
2 3 1
3 2 Y
1 3 Y
1 3


Input

1 3
100 50 1


Output

Impossible

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
int a[15][15],g[105][105];
int main()
{
	int n,m,i,j,x,y,k,l,o,O,cnt=200;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)for(j=1;j<=m;++j)scanf("%d",&a[i][j]);
	memset(g,-1,sizeof(g));
	for(i=1;i<=n;++i)for(j=1;j<=m;++j)
	{
		for(o=x=0;x<=100&&!o;++x)for(y=0;y<=100&&!o;++y)if(g[x][y]>=0&&g[x][y]+x*i+y*j==a[i][j])o=1;
		if(o)continue;
		for(x=0;x<=100&&!o;++x)for(y=0;y<=100&&!o;++y)if(g[x][y]<0&&x*i+y*j<=a[i][j])
		{
			for(O=0,k=1;k<=n&&!O;++k)for(l=1;l<=m&&!O;++l)if(x*k+y*l+a[i][j]-x*i-y*j<a[k][l])O=1;
			if(!O){g[x][y]=a[i][j]-x*i-y*j;o=1;++cnt;}
		}
		if(!o)return 0*puts("Impossible");
	}
	printf("Possible\n202 %d\n",cnt);
	for(i=1;i<=100;++i)printf("%d %d X\n",i,i+1);
	for(i=102;i<=201;++i)printf("%d %d Y\n",i,i+1);
	for(i=0;i<=100;++i)for(j=0;j<=100;++j)if(g[i][j]>=0)printf("%d %d %d\n",i+1,202-j,g[i][j]);
	puts("1 202");
}