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先倍增出 $i$使得 $2^i$時間時剛好有每個點能夠到 $m$

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read(){
	ll ans=0;
	char ch=getchar();
	while(!isdigit
 
(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans;
}
const int N=105;
int n;
ll m,val;
struct Matrix{
	ll a[N][N];
	inline void init(){memset(a,-1,sizeof(a));}
	friend inline Matrix operator*(const Matrix&a,const Matrix&b){
		Matrix ret;
		ret.init();
		for(register int i=1;i<=n;++i)for(register int k=1;k<=n;++k)if(~a.a[i][k])for(register int j=1;j<=n;++j)if(~b.a[k][j])ret.a[i][j]=max(ret.a[i][j],a.a[i][k]+b.a[k][j]);
		for(register int i=1;i<=n;++i)for(register int j=1;j<=n;++j)if(ret.a[i][j]>m)ret.a[i][j]=m;
		return ret;
	}
	inline bool check(){for(register int i=1;i<=n;++i)if(a[1][i]==m)return 1;return 0;}
}dis[105],mul,tmp;
int main(){
	for(register int tt=read();tt;--tt){
		n=read(),m=read();
		for(register int i=1;i<=n;++i)for(register int j=1;j<=n;++j){
			dis[0].a[i][j]=read();
			if(!dis[0].a[i][j])--dis[0].a[i][j];
		}
		int up=0;
		for(;;++up){
			dis[up+1]=dis[up]*dis[up];
			if(dis[up+1].check())break;
		}
		mul=dis[0];
		ll ans=1ll;
		for(register int i=up;~i;--i){
			tmp=mul*dis[i];
			if(!tmp.check()){
				for(register int j=1;j<=n;++j)for(register int k=1;k<=n;++k)mul.a[j][k]=tmp.a[j][k];
				ans+=1ll<<i;
			}
		}
		cout<<ans+1ll<<'\n';
	}
	return 0;
}