// Original C++ solution (slower):
#include <cstdio>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <queue>



using namespace std;



struct edge

{

    int v,w,nxt;

}e[400050];



int head[100050],ecnt;



void adde(int u,int v,int w)

{

    e[++ecnt].v = v;

    e[ecnt].w = w;

    e[ecnt].nxt = head[u];

    head[u] = ecnt;

}



int N,M;

int l[200050],r[200050];

int d[200050];

long long dist[100005];

queue<int> que;

bool vis[100050],inq[100050];



bool spfa(int st)

{

    que.push(st);

    inq[st] = true;

    dist[st] = 0;

    long long curmax = 0,curmin = 0;

    while(!que.empty())

    {

        int u = que.front();

        que.pop();

        for(int i = head[u];i;i = e[i].nxt)

        {

            int v = e[i].v;

            if(dist[v] == 12345678987654321)

            {

                dist[v] = dist[u] + e[i].w;

                curmax = max(curmax,dist[v]);

                curmin = min(curmin,dist[v]);

                if(curmax - curmin > 1000000000)

                    return false;

                vis[v] = 1;

                if(!inq[v])

                    que.push(v);

            }

            else if(dist[v] != dist[u] + e[i].w)

                return false;

        }

        inq[u] = false;

    }

    return true;

}



int main()

{

    for(int i = 1;i <= 100000; ++ i)

        dist[i] = 12345678987654321;

    scanf("%d%d",&N,&M);

    for(int i = 1;i <= M; ++ i)

    {

        scanf("%d%d%d",&l[i],&r[i],&d[i]);

        adde(l[i],r[i],d[i]);

        adde(r[i],l[i],-d[i]);

    }

    for(int i = 1;i <= N; ++ i)

    {

        if(!vis[i])

        {

            vis[i] = 1;bool l = spfa(i);

            if(!l)

            {

                printf("No\n");

                return 0;

            }

        }

    }

    printf("Yes\n");

    return 0;

}

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

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

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

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

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

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}

// Fastest known solution for this problem: