Constraints

* 1 ≤ |V| ≤ 100
* 0 ≤ |E| ≤ 9900
* -2 × 107 ≤ di ≤ 2 × 107
* There are no parallel edges
* There are no self-loops

Input

An edge-weighted graph G (V, E).


|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in G. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the cost of the i-th edge.

Output

If the graph contains a negative cycle (a cycle whose sum of edge costs is a negative value), print


NEGATIVE CYCLE


in a line.

Otherwise, print


D0,0 D0,1 ... D0,|V|-1
D1,0 D1,1 ... D1,|V|-1
:
D|V|-1,0 D1,1 ... D|V|-1,|V|-1


The output consists of |V| lines. For each ith line, print the cost of the shortest path from vertex i to each vertex j (j = 0, 1, ... |V|-1) respectively. If there is no path from vertex i to vertex j, print "INF". Print a space between the costs.

Examples

Input

4 6
0 1 1
0 2 5
1 2 2
1 3 4
2 3 1
3 2 7


Output

0 1 3 4
INF 0 2 3
INF INF 0 1
INF INF 7 0


Input

4 6
0 1 1
0 2 -5
1 2 2
1 3 4
2 3 1
3 2 7


Output

0 1 -5 -4
INF 0 2 3
INF INF 0 1
INF INF 7 0


Input

4 6
0 1 1
0 2 5
1 2 2
1 3 4
2 3 1
3 2 -7


Output

NEGATIVE CYCLE
import sys
input = sys.stdin.readline


def warshall_floid(graph: list) -> list:
    """warshall-floid法: 各頂点から各頂点への最短距離を求める
    計算量: O(V^3)
    """
    n = len(graph)
    INF = 10 ** 18
    dp = [[INF] * n for i in range(n)]
    for i in range(n):
        dp[i][i] = 0
    for i in range(n):
        for j, cost in graph[i]:
            dp[i][j] = cost

    for k in range(n):
        for i in range(n):
            for j in range(n):
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j])
    return dp
  

v, e = map(int, input().split())
info = [list(map(int, input().split())) for i in range(e)]
INF = 10 ** 18

graph = [[] for i in range(v)]
for a, b, cost in info:
    graph[a].append((b, cost))

dp = warshall_floid(graph)

is_negative = False
for i in range(v):
    if dp[i][i] < 0:
        is_negative = True

if is_negative:
    print("NEGATIVE CYCLE")
else:
    for res in dp:
        print(*[num if num < 10**12 else "INF" for num in res])
