Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 3
1 2 2
1 2 3

Sample Output 1

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

Sample Input 2

1 3
100 50 1

Sample Output 2

Impossible

Solution in C++ (Clang 10.0.0) (judge verdict: Runtime Error)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#include <iomanip>
#define vll vector<ll>
#define vvvl vector<vvl>
#define vvl vector<vector<ll>>
#define VV(a, b, c, d) vector<vector<d>>(a, vector<d>(b, c))
#define VVV(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));
#define re(c, b) for(ll c=0;c<b;c++)
#define all(obj) (obj).begin(), (obj).end()
typedef long long int ll;
typedef long double ld;
using namespace std;

typedef tuple<ll, ll, ll> tp;

int main(int argc, char const *argv[]) {
  // 不可能なものを削る
  ll a, b;std::cin >> a >> b;
  vvl d = VV(a+1, b+1, 0, ll);
  re(i, a) re(j, b) scanf("%lld", &d[i+1][j+1]);
  ll INF = 1000000000;
  vvl cost = VV(a+1, b+1, INF, ll);

  vector<tp> use;
  for(ll i=0;i<=99;i++){
    for(ll j=0;j<=99;j++){
      for(ll k=0;k<=100;k++){
        //i, j, kの時のコストが下回ったら絶対に使えない
        for(ll A=1;A<=a;A++){
          for(int B=1;B<=b;B++){
            if(i*A+j*B+k<d[A][B]) goto nxt;
          }
        }
        use.push_back(tp(i, j, k));//採用
        for(int A=1;A<=a;A++){
          for(int B=1;B<=b;B++){
            cost[A][B] = min(cost[A][B], i*A+j*B+k);
          }
        }
        nxt:;
      }
    }
  }

  for(int i=1;i<=a;i++){
    for(int j=1;j<=b;j++){
      if(cost[i][j]!=d[i][j]){
        std::cout << "Impossible" << '\n';
        return 0;
      }
    }
  }
  return 1;
  std::cout << "Possible" << '\n';
  vvl e;
  vvl ok = VV(301, 301, 0, ll);

  for(int i=1;i<100;i++){
    ok[i][i+1] = 1;
    e.push_back({i, i+1, -1});
  }
  for(int i=1;i<100;i++){
    ok[i+200][i+201] = 1;
    e.push_back({i+200, i+201, -2});
  }

  for(auto v:use){
    ll i = get<0>(v);
    ll j = get<1>(v);
    ll k = get<2>(v);
    ll x = 0;
  }

  //------output area-----
  std::cout << "300 " << e.size() << '\n';


  std::cout << "1 300" << '\n';
  return 0;
}