// Original C++ solution (slower):
//include

//------------------------------------------

#include <bits/stdc++.h>

using namespace std;



//typedef

//------------------------------------------

typedef vector<int> VI;

typedef vector<VI> VVI;

typedef vector<string> VS;

typedef pair<int, int> PII;

typedef long long LL;



//container util

//------------------------------------------

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

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

#define PB push_back

#define MP make_pair

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

#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)

#define EXIST(s,e) ((s).find(e)!=(s).end())

#define SORT(c) sort((c).begin(),(c).end())



//constant

//--------------------------------------------

const double EPS = 1e-10;

const double PI  = acos(-1.0);

const int MOD = 1000000007;

const int INF = 2147483647;



//debug

#define dump(x)  cerr << #x << " = " << (x) << endl;

#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;



//myfuncs

//素数判定

template<class T> inline T isPrime(T x) {

    bool ret = true;

    if (x < 2) {

        ret = false;

    }

    for (LL i = 2; i <= sqrt(x); i++) {

        if (x%i == 0) {

            ret = false;

        }

    }

    return ret;

}



//Union FInd

struct UnionFind {

    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2



    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化

        for(int i = 0; i < N; i++) par[i] = i;

    }



    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}

        if (par[x] == x) return x;

        return par[x] = root(par[x]);

    }



    void unite(int x, int y) { // xとyの木を併合

        int rx = root(x); //xの根をrx

        int ry = root(y); //yの根をry

        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま

        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける

    }



    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す

        int rx = root(x);

        int ry = root(y);

        return rx == ry;

    }

};



//main code



const int NMAX = 100000;

// VVI x(NMAX,VI(NMAX,0));    10^5 * 10^5はオーバーフローっぽい

vector<vector<pair<int,int>>> x(NMAX);

vector<bool> visited(NMAX,false);

vector<int> sum_cost(NMAX,0);





void dfs(int v,int n){

    visited[v] = true;



    for (int i = 0; i < x[v].size(); i++) {

        int next_v = x[v][i].first;

        int cost = x[v][i].second;

        // すでにコストが入力されている時、トータルコストに矛盾が生じないかの検証

        if (sum_cost[next_v] != 0) {

            if (sum_cost[next_v] != sum_cost[v] + cost) {

                cout << "No" << endl;

                exit(0);

            }

        }

        // 矛盾がなければ探索続行

        else if (!visited[next_v]) {

            sum_cost[next_v] = sum_cost[v] + cost;

            dfs(x[v][i].first,n);

        }



    }

}





int main(int argc, char const* argv[])

{

    int n,m;

    cin >> n >> m;

    set<int> all_v;  // 複数のグラフに分かれた時に対応するため

    if (m == 0) {

        cout << "Yes" << endl;

        return 0;

    }

    int l,r,d;

    for (int i = 0; i < m; i++) {

        cin >> l >> r >> d;

        x[l-1].push_back(make_pair(r-1,d));

        x[r-1].push_back(make_pair(l-1,-d));

        all_v.insert(l-1);

        all_v.insert(r-1);

    }



    // 訪れる必要がある頂点に対して、まだ到達していなければdfs実行

    for (auto&& v : all_v) {

        if (!visited[v]) {

            dfs(v,n);

        } 

    }



    // 矛盾なくdfs抜けたらyes

    cout << "Yes" << endl;

    return 0;

}


// Optimized solution to the same problem (faster):
//include

//------------------------------------------

#include <bits/stdc++.h>

using namespace std;



//typedef

//------------------------------------------

using LL = int64_t;

using VL = vector<LL>;

using VVL = vector<VL>;

using VS = vector<string>;

using PLL = pair<LL, LL>;



//container util

//------------------------------------------

#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)

#define rwhole(f,x,...) ([&](decltype((x)) whole) { return (f)(rbegin(whole), rend(whole), ## __VA_ARGS__); })(x)

#define EACH(i,c) for(decltype((c).begin()) i=(c).begin(); i!=(c).end(); ++i)

#define EXIST(s,e) ((s).find(e)!=(s).end())

#define ALL(x) ::std::begin(x), ::std::end(x)

#define RALL(x) ::std::rbegin(x), ::std::rend(x)

#define SORT(c) whole(sort, c)

#define RSORT(c) rwhole(sort, c)



//constant

//--------------------------------------------

constexpr double EPS = 1e-10;

constexpr double PI  = 3.14159265358979323846;

constexpr int MOD = 1000000007;



// grid

//--------------------------------------------

VL dx = {0, 1, 0, -1};

VL dy = {1, 0, -1, 0};

VL dx2 = {-1, 0, 1, -1, 1, -1, 0, 1};

VL dy2 = {-1, -1, -1, 0, 0, 1, 1, 1};



//debug

//--------------------------------------------

#define dump(x)  cerr << #x << " = " << (x) << endl;

#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;



//IO accelerate

//--------------------------------------------

struct InitIO {

    InitIO() {

        cin.tie(nullptr);

        ios_base::sync_with_stdio(false);

        cout << fixed << setprecision(30);

    }

} init_io;



//template

//--------------------------------------------

// declaretion

template<typename T> istream& operator >>(istream& is, vector<T>& vec);

template<typename T1, typename T2> ostream& operator <<(ostream& os, const pair<T1, T2>& p);

template<typename T> ostream& operator <<(ostream& os, const vector<T>& vec);

template<typename T> ostream& operator <<(ostream& os, const vector<vector<T>>& vv);

template<typename T> vector<T> make_v(size_t a);

template<typename T,typename... Ts> auto make_v(size_t a,Ts... ts);

template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type fill_v(T &t,const V &v);

template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type fill_v(T &t,const V &v);



// implementation

template<typename T>

istream& operator >>(istream& is, vector<T>& vec) {

	for(T& x: vec) is >> x;

	return is;

}

template<typename T1, typename T2>

ostream& operator <<(ostream& os, const pair<T1, T2>& p) {

	os << p.first << "," << p.second;

	return os;

}

template<typename T>

ostream& operator <<(ostream& os, const vector<T>& vec) {

	for(int i=0; i<vec.size(); i++){

		os << vec[i] << ( i+1 == vec.size() ? "" : "\t" );

	}

	return os;

}

template<typename T>

ostream& operator <<(ostream& s, const vector<vector<T>>& vv) {

	for (int i = 0; i < vv.size(); ++i) {

		s << vv[i] << endl;

	}

	return s;

}



// 多重vector

// auto dp=make_v<int>(4,h,w) みたいに使える

template<typename T>

vector<T> make_v(size_t a){return vector<T>(a);}



template<typename T,typename... Ts>

auto make_v(size_t a,Ts... ts){

	return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));

}



// 多重vectorのためのfill

// fill_v(dp,0) みたいに使える

template<typename T,typename V>

typename enable_if<is_class<T>::value==0>::type

fill_v(T &t,const V &v){t=v;}



template<typename T,typename V>

typename enable_if<is_class<T>::value!=0>::type

fill_v(T &t,const V &v){

	for(auto &e:t) fill_v(e,v);

}



template<class T,class U> void chmax(T&a,U b){if(a<b)a=b;}

template<class T,class U> void chmin(T&a,U b){if(b<a)a=b;}

template<typename T> T gcd(T a, T b) { return b?gcd(b,a%b):a;}

template<typename T> T lcm(T a, T b) { return a/gcd(a,b)*b;}



// auto mod int

//

// https://youtu.be/L8grWxBlIZ4?t=9858

// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize

// https://youtu.be/8uowVvQ_-Mo?t=1329 : division

// const int MOD = 1000000007;

struct mint {

	LL x; // typedef long long LL;

	constexpr mint(LL x=0):x((x%MOD+MOD)%MOD){}

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

	constexpr mint& operator+=(const mint& a) {

		if ((this->x += a.x) >= MOD) this->x -= MOD;

		return *this;

	}

	constexpr mint& operator-=(const mint& a) {

		if ((this->x += MOD-a.x) >= MOD) this->x -= MOD;

		return *this;

	}

	constexpr mint& operator*=(const mint& a) {

		(this->x *= a.x) %= MOD;

		return *this;

	}

	constexpr mint operator+(const mint& a) const {

		mint res(*this);

		return res+=a;

	}

	constexpr mint operator-(const mint& a) const {

		mint res(*this);

		return res-=a;

	}

	constexpr mint operator*(const mint& a) const {

		mint res(*this);

		return res*=a;

	}

	constexpr 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

	constexpr mint inv() const {

		return pow(MOD-2);

	}

	constexpr mint& operator/=(const mint& a) {

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

	}

	constexpr mint operator/(const mint& a) const {

		mint res(*this);

		return res/=a;

	}



	constexpr bool operator==(const mint& a) const {

		return this->x == a.x;

	}

	constexpr bool operator!=(const mint& a) const {

		return this->x != a.x;

	}

	constexpr bool operator<(const mint& a) const {

		return this->x < a.x;

	}



	friend istream& operator>>(istream& is, mint& m) {

		return is >> m.x;

	}

	friend ostream& operator<<(ostream& os, const mint& m) {

		return os << m.x;

	}

};

typedef vector<mint> VM;

typedef vector<VM> VVM;



//main code

struct edge {

	LL to, cost;

	edge(LL to, LL cost): to(to) ,cost(cost) {}

};



int main(int argc, char *argv[])

{

	LL n,m;

	cin >> n >> m;

	VL iti(n, INT64_MIN);

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

	for (int i = 0; i < m; i++) {

		int l,r,d;

		cin >> l >> r >> d;

		l--;r--;

		g[l].emplace_back(edge(r,d));

		g[r].emplace_back(edge(l,-d));

	}

	vector<bool> visited(n,false);

	auto dfs = [&](auto&& dfs, int v, int par) -> void{

		visited[v] = true;

		for (auto e : g[v]) {

			if (e.to == par) continue;

			if (iti[e.to] == INT64_MIN) {

				iti[e.to] = iti[v] + e.cost;

				dfs(dfs, e.to, v);

			} else {

				if (iti[e.to] != iti[v] + e.cost) {

					cout << "No" << endl;

					exit(0);

				}

			}

		}

		return;

	};

	for (int i = 0; i < n; i++) {

		if (!visited[i]) {

			iti[i] = 0;

			dfs(dfs, i,-1);

		}

	}

	cout << "Yes" << endl;

	return 0;

}

// Fastest known solution for this problem: