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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;

int main(){
    int v,e;
    cin >> v >> e;
    vector<vector<ll>> dist(v,vector<ll>(v,LINF));
    rep(i,v) dist[i][i] = 0;
    rep(i,e){
        ll s,t,d;
        cin >> s >> t >> d;
        dist[s][t] = d;
    }

    rep(k,v)rep(i,v)rep(j,v){
        if(dist[i][k]!=LINF&&dist[k][j]!=LINF) dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j]);
    }

    rep(i,v){
        if(dist[i][i]<0){
            cout << "NEGATIVE CYCLE" << endl;
            return 0;
        }
    }

    rep(i,v)rep(j,v){
        if(dist[i][j]>=LINF) cout << "INF" <<  (j==v-1?'\n':' ');
        else cout << dist[i][j] <<  (j==v-1?'\n':' ');
    }

    return 0;
}
