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 <algorithm>
#include <cstdio>
const int N = 12;
const int x = 0x7fffffff;
const int y = 0x7ffffffe;
int d[N][N];
bool v[N][N];
int X[1000], Y[1000], S, T, cnt, m;
int from[1000000], to[1000000], cost[1000000];
inline void addEdge(int x, int y, int z) {
  cost[m] = z;
  from[m] = x;
  to[m++] = y;
}
int main() {
  int A, B;
  scanf("%d%d", &A, &B);
  int xx = 0, yy = 0;
  for (int i = 1; i <= A; ++i)
    for (int j = 1; j <= B; ++j) {
      scanf("%d", &d[i][j]);
      xx = std::max(xx, d[i][j] / i);
      yy = std::max(yy, d[i][j] / j);
      v[i][j] = 0;
    }
  cnt = 0;
  m = 0;
  for (int i = 0; i <= xx; ++i) X[i] = cnt++;
  for (int i = 0; i <= yy; ++i) Y[i] = cnt++;
  S = X[0];
  T = Y[0];
  for (int i = 0; i < xx; ++i) addEdge(X[i], X[i + 1], x);
  for (int i = 0; i < yy; ++i) addEdge(Y[i + 1], Y[i], y);
  for (int i = 0; i <= xx; ++i)
    for (int j = 0; j <= yy; ++j) {
      int ans = 0;
      for (int a = 1; a <= A; ++a)
        for (int b = 1; b <= B; ++b)
          ans = std::max(ans, d[a][b] - i * a - j * b);
      if (ans < 0 || ans > 100) continue;
      addEdge(X[i], Y[j], ans);
      for (int a = 1; a <= A; ++a)
        for (int b = 1; b <= B; ++b)
          if (d[a][b] == i * a + j * b + ans)
            v[a][b] = true;
    }
  for (int a = 1; a <= A; ++a)
    for (int b = 1; b <= B; ++b)
      if (!v[a][b]) return puts("Impossible") & 0;
  puts("Possible");
  printf("%d %d\n", cnt, m);
  for (int i = 0; i < m; ++i) {
    printf("%d %d ", 1 + from[i], 1 + to[i]);
    if (cost[i] == x) printf("X");
    else if (cost[i] == y) printf("Y");
    else printf("%d", cost[i]);
    printf("\n");
  }
  printf("%d %d\n", 1 + S, 1 + T);
  return 0;
}