#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define vi vector<ll>
#define f(i,a,b) for(ll i=a;i<b;i++)
#define fa(e,l) for(auto e:l)
#define all(arr) arr.begin(),arr.end()
// int dfs(ll i,vi& visited,vector<vi>& adj,vi& trip,ll par){
// fa(e,adj[i]){
// if(e==par){continue;}
// if(visited[e]==0){return e;}
// else{
// visited[e]=0;
// ll a=dfs(e,visited,adj,trip,i);
// if(a!=-1){return a;}
// }
// }
// return -1;
// }
// int bfs(){return 0;}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
ll n,k,m=0,q;
ll a,b,c,d=0,f,g;
ll x,y,z;
ll w;
ll p,r,t,u,l;
ll MOD=1000000007;
ll INF=1e16;
ll T;
ll t2,t1,x1,x2,y1,y2;
// cin>>T;
T=1;
f(I,0,T){
cin>>n>>m;
// string s;
// cin>>s;
ll ans=0;
vi arr(n,-1);
vector<tuple<ll,ll,ll>> q;
f(i,0,m){
cin>>a>>b>>c;
q.pb({c,a,b});
}
sort(all(q));
f(i,0,m){
tie(c,a,b)=q[i];
arr[a-1]=max(c,arr[a-1]);
arr[b-1]=max(c,arr[b-1]);
}
fa(e,arr){
if(e==-1){cout<<"1 ";}
cout<<e<<" ";
}
}
}
/*
5 5
For odd pi, put pi+1/2 into winning class
For even pi, put pi/2+1 into winning class
If possible, nice
If total sum of x+y is >=sum of all pi
And if x-y>=total wins of A-total wins of B, then true
2x2
x3x
111x
xxx1
22
xx
xxxx Inchresting
2222
26
x
23
xx
kx+1
Addition is constant
It doesnt change
So at each new gate +x
x will be added no matter the amount of people
*2 or *3
So lets say
For each gate
2 3 2 3 2
3 3 3 3 3
4
2
1 x a c e g
1 y b d f h
1*(rest of multiples top - 5)
1*(rest of multiples bottom -5)
In the first case, here is one possible way to play this game optimally.
Initially, we have l=1
person in the left lane and r=1
person in the right lane.
After passing through the first pair of gates, we gain 4
people from the left gate and 1⋅(2−1)=1
person from the right gate, for a total of 4+1=5
people. We allocate 2
people to the left lane and 3
people to the right lane. This results in l=1+2=3
people in the left lane and r=1+3=4
people in the right lane.
After passing through the second pair of gates, we gain 3⋅(3−1)=6
people from the left gate and 4⋅(3−1)=8
people from the right gate, for a total of 6+8=14
people. We allocate 7
people to the left lane and 7
people to the right lane. This results in l=3+7=10
people in the left lane and r=4+7=11
people in the right lane.
After passing through the last pair of gates, we gain 7
people from the left gate and 4
people from the right gate, for a total of 7+4=11
people. We allocate 6
people to the left lane and 5
people to the right lane. This results in l=10+6=16
people in the left lane and r=11+5=16
people in the right lane.
At the end, the total number of people is 16+16=32
.
*/
