// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

#define rep(i, n) for (ll i = 0; i < n; i++)

#define FOR(i, a, b) for (ll i = a; i < b; i++)

#define len(v) ll(v.size())

#define fi first

#define se second



template <class T>

void cout_vec(const vector<T> &vec){

  for(auto itr:vec) cout<<itr<<' ';

  cout<<endl;

}



typedef pair<ll,ll> P;

const ll mod=1e9+7;

const int inf=-1e9;



struct edge{int to,cost;};



int n,m;

vector<vector<edge>> g(100010);

vector<int> d(100010,inf);



bool dfs(int v,int now){

  if(d[v]!=inf){

    if(d[v]!=now) return false;

    else return true;

  }

  d[v]=now;

  bool ret=true;

  for(auto itr:g[v]){

    ret=dfs(itr.to,now+itr.cost);

    if(!ret) break;

  }

  return ret;

}



int main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  cin>>n>>m;

  rep(i,m){

    int l,r,d;

    cin>>l>>r>>d;

    l--,r--;

    g[l].push_back({r,d});

    g[r].push_back({l,-d});

  }

  bool flag=true;

  rep(i,n){

    if(d[i]==inf){

      flag=dfs(i,1e5);

    }

    if(!flag) break;

  }

  if(flag) cout<<"Yes"<<endl;

  else cout<<"No"<<endl;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

typedef pair<ll,ll> P;

typedef pair<int,int> Pi;

#define rep(i,n) for(ll i=0;i<n;i++)

#define FOR(i,a,b) for(ll i=a;i<b;i++)

#define fi first

#define se second

#define endl "\n"



template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}

template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}

template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}

template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){int len=d.size();rep(i,len){s<<d[i];if(i<len-1) s<<" ";}return s;}

template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){int len=d.size();rep(i,len){s<<d[i]<<endl;}return s;}

template<typename T> ostream& operator<<(ostream& s,const set<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){s<<"{"<<endl;for(auto itr=m.begin();itr!=m.end();++itr){s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;}s<<"}"<<endl;return s;}



const ll mod=1e9+7;

const ll inf=1e17;

const int INF=1e9;

const double PI=acos(-1);

const double EPS=1e-10;



struct edge{int to,cost;};



int main(){

    cin.tie(0);ios::sync_with_stdio(false);

    int n,m;

    cin>>n>>m;

    vector<vector<edge>> g(n);

    vector<int> l(m),r(m),d(m);

    rep(i,m){

        cin>>l[i]>>r[i]>>d[i];

        l[i]--,r[i]--;

        g[l[i]].push_back({r[i],d[i]});

        g[r[i]].push_back({l[i],-d[i]});

    }

    vector<int> ans(n,-INF);

    int id=0;

    auto bfs=[&](int v){

        queue<int> que;

        que.push(v);

        ans[v]=0;

        while(!que.empty()){

            int from=que.front();

            que.pop();

            for(auto to:g[from]){

                if(ans[to.to]!=-INF) continue;

                ans[to.to]=ans[from]+to.cost;

                que.push(to.to);

            }

        }

    };

    rep(i,n){

        if(ans[i]==-INF){

            bfs(i);

        }

    }

    rep(i,m){

        if(ans[r[i]]-ans[l[i]]!=d[i]){

            cout<<"No"<<endl;

            return 0;

        }

    }

    cout<<"Yes"<<endl;

}

// Fastest known solution for this problem: