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

using namespace std;



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

#define ALL(a) (a).begin(), (a).end()

#define rALL(a) (a).rbegin(), (a).rend()

#define SZ(x) ((int)(x).size())

using ll = long long;

using ull = unsigned long long;

using P = pair<int, int>;

using llP = pair<ll, ll>;

using DoP = pair<double, double>;

const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};

const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};

const int INF = 1 << 30;

const ll INF64 = 1ll << 55;

const int mod = 1000000007;

//const int mod = 998244353;



template <class T>inline bool chmin(T &a, const T b)

{

  if (a > b)

  {

    a = b;

    return true;

  }

  return false;

}

template <class T>inline bool chmax(T &a, const T b)

{

  if (a < b)

  {

    a = b;

    return true;

  }

  return false;

}

template <class T>inline void line_out(const vector<T> vec) {

  int n = SZ(vec);

  rep(i, n) {

    cout << vec[i];

    if(i < n-1) cout << " ";

  }

  cout << endl;

  return;

}



struct mint

{

  ll x; // typedef long long ll;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint &operator+=(const mint a)

  {

    if ((x += a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator-=(const mint a)

  {

    if ((x += mod - a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator*=(const mint a)

  {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const

  {

    mint res(*this);

    return res += a;

  }

  mint operator-(const mint a) const

  {

    mint res(*this);

    return res -= a;

  }

  mint &operator++()

  {

    if ((x += 1) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator--()

  {

    if ((x += mod - 1) >= mod)

      x -= mod;

    return *this;

  }

  mint operator*(const mint a) const

  {

    mint res(*this);

    return res *= a;

  }

  mint pow(ll t) const

  {

    if (!t)

      return 1;

    mint a = pow(t >> 1);

    a *= a;

    if (t & 1)

      a *= *this;

    return a;

  }



  // for prime mod

  mint inv() const

  {

    return pow(mod - 2);

  }

  mint &operator/=(const mint a)

  {

    return (*this) *= a.inv();

  }

  mint operator/(const mint a) const

  {

    mint res(*this);

    return res /= a;

  }

};



struct edge {int to; ll cost;};

vector<vector<edge>> g(100000+1);

vector<ll> dist(100000+1, INF64);



void dfs(int now) {

  for(auto next : g[now]) {

    if(dist[next.to] == INF64) {

      dist[next.to] = dist[now]+next.cost;

      dfs(next.to);

    }

  }

}



int main()

{

  int n, m; cin >> n >> m;

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

  rep(i, m) {

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

    edge x = {r[i], d[i]};

    edge y = {l[i], -d[i]};

    g[l[i]].push_back(x);

    g[r[i]].push_back(y);

  }



  for(int i = 1; i <= n; ++i) {

    if(dist[i] == INF64) {

      dist[i] = 0;

      dfs(i);

    }

  }

  rep(i, m) {

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

      cout << "No" << endl;

      return 0;

    }

  }

  cout << "Yes" << endl;

}

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

using namespace std;



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

#define ALL(a) (a).begin(), (a).end()

#define rALL(a) (a).rbegin(), (a).rend()

#define SZ(x) ((int)(x).size())

using ll = long long;

using ull = unsigned long long;

using P = pair<int, int>;

using llP = pair<ll, ll>;

using DoP = pair<double, double>;

const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};

const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};

const int INF = 1 << 28;

const ll INF64 = 1ll << 55;

const int mod = 1000000007;

//const int mod = 998244353;



template <class T>inline bool chmin(T &a, const T b)

{

  if (a > b)

  {

    a = b;

    return true;

  }

  return false;

}

template <class T>inline bool chmax(T &a, const T b)

{

  if (a < b)

  {

    a = b;

    return true;

  }

  return false;

}

template <class T>inline void line_out(const vector<T> vec) {

  int n = SZ(vec);

  rep(i, n) {

    cout << vec[i];

    if(i < n-1) cout << " ";

  }

  cout << endl;

  return;

}



struct mint

{

  ll x; // typedef long long ll;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint &operator+=(const mint a)

  {

    if ((x += a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator-=(const mint a)

  {

    if ((x += mod - a.x) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator*=(const mint a)

  {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const

  {

    mint res(*this);

    return res += a;

  }

  mint operator-(const mint a) const

  {

    mint res(*this);

    return res -= a;

  }

  mint &operator++()

  {

    if ((x += 1) >= mod)

      x -= mod;

    return *this;

  }

  mint &operator--()

  {

    if ((x += mod - 1) >= mod)

      x -= mod;

    return *this;

  }

  mint operator*(const mint a) const

  {

    mint res(*this);

    return res *= a;

  }

  mint pow(ll t) const

  {

    if (!t)

      return 1;

    mint a = pow(t >> 1);

    a *= a;

    if (t & 1)

      a *= *this;

    return a;

  }



  // for prime mod

  mint inv() const

  {

    return pow(mod - 2);

  }

  mint &operator/=(const mint a)

  {

    return (*this) *= a.inv();

  }

  mint operator/(const mint a) const

  {

    mint res(*this);

    return res /= a;

  }

};

struct edge {int to; ll cost;};





int main()

{

  int n, m;

  cin >> n >> m;

  vector<vector<edge>> g(n+1);

  int init;

  rep(i, m) {

    int l, r, d;

    cin >> l >> r >> d;

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

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

  }

  vector<ll> node(n+1, INF64);

  queue<int> q;



  for(int s = 1; s <= n; ++s) {

    if(node[s] != INF64) continue;



    node[s] = 0;

    q.push(s);



    while(!q.empty()) {

      int now = q.front(); q.pop();



      for(auto next : g[now]) {

        if(node[next.to] == INF64) {

          node[next.to] = node[now]+next.cost;

          q.push(next.to);

        }

        else if(node[next.to] != node[now]+next.cost) {

          cout << "No" << endl;

          return 0;

        }

      }



    }

  }



  cout << "Yes" << endl;

}

// Fastest known solution for this problem: