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>

inline int Min(int a,int b){return a<b?a:b;}
inline int Max(int a,int b){return a>b?a:b;}

const int N=305,Inf=0x3f3f3f3f;
int n,m,f[N][N],d[N][N];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            scanf("%d",&d[i][j]);
    for(int i=0;i<=100;++i)
        for(int j=0;j<=100;++j)
            for(int x=1;x<=n;++x)
                for(int y=1;y<=m;++y)
                    f[i][j]=Max(f[i][j],d[x][y]-i*x-j*y);
    for(int x=1;x<=n;++x)
        for(int y=1;y<=m;++y)
        {
            int Now=Inf;
            for(int i=0;i<=100;++i)
                for(int j=0;j<=100;++j)
                    Now=Min(Now,f[i][j]+i*x+j*y);
            if(Now!=d[x][y])return puts("Impossible"),0;
        }
    puts("Possible\n202 10401");
    for(int i=1;i<=100;++i)printf("%d %d X\n",i,i+1);
    for(int i=102;i<=201;++i)printf("%d %d Y\n",i,i+1);
    for(int i=0;i<=100;++i)
        for(int j=0;j<=100;++j)
            printf("%d %d %d\n",i+1,202-j,f[i][j]);
    return puts("1 202"),0;
}