/*
Baltic 2020 Graph
- N.B. The graph isn't necessarily connected, so apply the following on all connected components
- Let node 1 have value X
- Then do a DFS
- We know the value of each node in terms of X from the previously visited node and edge
- This means we get a bunch of linear equations (mx + c) for each node
- If there is more than 1 linear equation for any node, we can either find X or show that a labelling
is not possible
- If all nodes only have 1 linear equation though, then we must find X to minimize the sum of their
absolute values
- Notice how we can simply sort by (-m * c) and take the median in this case, since abs(m) = 1
- Complexity: O(N log N)
*/
#include <bits/stdc++.h>
#define FOR(i, x, y) for (ll i = x; i < y; i++)
typedef long long ll;
using namespace std;
pair<ll, ll> labels[100001];
vector<pair<ll, ll>> graph[100001];
ll twice_x, ans[100001];
vector<ll> processed;
bool know_x = false, visited[100001];
bool incompatible(pair<ll, ll> A, pair<ll, ll> B) {
if (know_x)
return (A.first * twice_x + 2 * A.second !=
B.first * twice_x + 2 * B.second);
return (A.first == B.first && A.second != B.second);
}
void calc_x(pair<ll, ll> A, pair<ll, ll> B) {
if (A.first == -1) swap(A, B);
twice_x = B.second - A.second;
know_x = true;
}
void dfs(ll node = 1) {
visited[node] = true;
processed.push_back(node);
for (pair<ll, ll> i : graph[node]) {
pair<ll, ll> assign = {-labels[node].first,
i.second - labels[node].second};
if (!visited[i.first]) {
labels[i.first] = assign;
dfs(i.first);
} else {
if (labels[i.first] == assign) continue;
if (incompatible(labels[i.first], assign)) {
cout << "NO\n";
exit(0);
}
calc_x(labels[i.first], assign);
}
}
}
void assign_opt_x() {
vector<int> salient_points;
for (int i : processed)
salient_points.push_back(-labels[i].first * labels[i].second);
sort(salient_points.begin(), salient_points.end());
twice_x = 2 * salient_points[salient_points.size() / 2];
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
ll n, m;
cin >> n >> m;
while (m--) {
ll u, v, w;
cin >> u >> v >> w;
graph[u].push_back({v, w});
graph[v].push_back({u, w});
}
FOR(i, 1, n + 1) {
if (!visited[i]) {
processed.clear();
know_x = false;
labels[i] = {1, 0};
dfs(i);
if (!know_x) assign_opt_x();
for (int i : processed)
ans[i] = labels[i].first * twice_x + 2 * labels[i].second;
}
}
cout << "YES\n";
FOR(i, 1, n + 1) cout << ans[i] / 2.0 << " \n"[i == n];
return 0;
}
