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
#include "bits/stdc++.h"

using namespace std;

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<vector<long>> dp(n, vector<long>(n, 1e15));
    while (m--)
    {
        int s, t;
        long d;
        cin >> s >> t >> d;
        dp[s][t] = d;
    }
    for (int k = 0; k < n; k++)
    {
        dp[k][k] = 0;
    }
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
            }
    for (int k = 0; k < n; k++)
    {
        if (dp[k][k] < 0)
        {
            cout << "NEGATIVE CYCLE\n";
            return;
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (dp[i][j] >= 1e12)
            {
                cout << "INF";
            }
            else
            {
                cout << dp[i][j];
            }
            cout << " \n"[j == n - 1];
        }
    }
}

int main()
{
    solve();
    return 0;
}

