/*
* Autor: Jerzy Olkowski
* Rozwiązanie wzorcowe
* Złożoność: O(nlogn + m)
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 100 * 1000 + 7;
const int I = 1000 * 1000 * 1000;
vector<pair<int, int>> ed[N];
pair<int, int> val[N];
vector<int> cur;
long double answer[N];
void DFS(int v)
{
cur.push_back(v);
for(int i = 0; i < (int)ed[v].size(); ++i)
{
if(val[ed[v][i].first].first != 0) continue;
val[ed[v][i].first].first = -val[v].first;
val[ed[v][i].first].second = ed[v][i].second - val[v].second;
DFS(ed[v][i].first);
}
}
bool Do()
{
bool czy = 1;
vector<int> pos;
long double x = -I - 10;
int ts = -I * 2;
for(int i = 0; i < (int)cur.size(); ++i)
{
int v = cur[i];
pos.push_back(val[v].first * val[v].second * -1);
for(int j = 0; j < (int)ed[v].size(); ++j)
{
int ne = ed[v][j].first;
if(val[v].first == val[ne].first)
{
if(ts > -I * 2 && ts != (ed[v][j].second - val[v].second - val[ne].second) * val[v].first)
czy = 0;
ts = (ed[v][j].second - val[v].second - val[ne].second) * val[v].first;
}
else if(val[v].second + val[ne].second != ed[v][j].second)
czy = 0;
}
}
sort(pos.begin(), pos.end());
if(ts == -I * 2)
x = pos[(int)pos.size() / 2];
else
x = (long double)ts / (long double)2;
for(int i = 0; i < (int)cur.size(); ++i)
{
int v = cur[i];
answer[v] = x * (long double)val[v].first + (long double)val[v].second;
}
cur.clear();
return czy;
}
void Wynik(int &n)
{
bool czy = 1;
for(int i = 1; i <= n; ++i)
{
if(val[i].first != 0) continue;
val[i].first = 1;
DFS(i);
czy = (czy & Do());
}
if(czy == 0)
{
cout << "NO\n";
return;
}
cout << "YES\n";
cout << fixed << setprecision(1);
for(int i = 1; i <= n; ++i){
cout << answer[i];
if (i != n) cout << ' ';
}
cout << "\n";
}
void WczytajLiczby(int &n)
{
int m, a, b, c;
cin >> n >> m;
for(int i = 1; i <= m; ++i)
{
cin >> a >> b >> c;
ed[a].push_back(make_pair(b, c));
ed[b].push_back(make_pair(a, c));
}
}
void Gra()
{
int n;
WczytajLiczby(n);
Wynik(n);
}
int main ()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
Gra();
return 0;
}
