// Slower C++ solution:
#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <unistd.h>

#include <string.h>

#include <algorithm>

#include <limits.h>

#include <vector>

#include <map>

#include <stack>

#include <queue>

#include <math.h>

#define ll long long

#define MSET(x) memset((x), 0, sizeof((x)))

#define MMSET(x) memset((x), -1, sizeof((x)))

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

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

#define repD(i, a, b) for ( ll (i) = a; (i) >= (b) ; (i) -- )

#define fi first

#define se second

#define pb push_back

#define mfind(m, item) ((m).find((item)) != (m).end())

#define mp(a,b) make_pair((a),(b))

#define mt(a,b,c) make_pair(make_pair((a),(b)), (c))

#define tfi first.first

#define tse first.second

#define tth second

#define prll cout << endl

#define tri pair<pair<ll, ll>, ll>

using namespace std;



int n,m;

int cnt[100002], ans[100002];



void find(int x){

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

    find(ans[x]);

    cnt[x] += cnt[ans[x]], ans[x] = ans[ans[x]];

}



bool has = false;



int main(){

    while ( cin >> n >> m ) {

        MSET(cnt); MSET(ans); has = false;

        for ( int i = 1 ; i <= n ; ++ i ) ans[i] = i;

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

            int l, r, d;

            scanf("%d %d %d", &l, &r, &d);

            find(l); find(r);

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

                if ( cnt[l] + d != cnt[r] ) {

                    has = true;

                }

            }

            else {

                cnt[ans[r]] = cnt[l] + d - cnt[r];

                ans[ans[r]] = ans[l];

            }

        }

        cout << (has ? "No" : "Yes" ) << endl;

    }

}


// Faster C++ solution for the same problem (measured speedup: 1.7664655807241882x):
#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/trie_policy.hpp>

#include <ext/rope>



//#pragma 03

//#pragma GCC optimize("Ofast")

//#pragma GCC optimization("unroll-loops, no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2")



#define ll long long

#define ld long double

#define fi first

#define se second

#define sz(x) x.size()

#define mp make_pair

#define pb push_back

#define eb emplace_back

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)

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

#define fastio() ios_base::sync_with_stdio(NULL); cin.tie(0); cout.tie(0)

//#define int long long



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());



typedef vector<int> vi;

typedef vector<vi> vvi;

typedef pair<int, int> ii;

typedef vector<pair<int, int> > vii;

typedef vector<ll> vl;

typedef vector<vl> vvl;

typedef pair<ll, ll> pll;

typedef vector<pll> vll;



inline void EXECUTE_TIME() {cout << endl << setprecision(10) << fixed << "CPU Time: " << clock() / double(CLOCKS_PER_SEC) << " sec(s)\n";}



template <class T> inline T RD(T &a) {T x; cin >> x; return a = x;}

template <class T> inline T gcd(T a, T b) {if (b) return gcd(b, a % b); else return a;}

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

template<typename T> using ordered_set=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef trie<string, null_type, trie_string_access_traits<>, pat_trie_tag, trie_prefix_search_node_update> pref_trie;



const int N = 1e5 + 1;



vector<vii> adj;

bool check[N];

int dist[N];

bool FAIL;



void dfs(int u) {

    if (!check[u]) check[u] = true;

    for (ii v : adj[u]) {

        if (!check[v.fi]) {

            dist[v.fi] = dist[u] - v.se;

            dfs(v.fi);

        } else {

            if (dist[u] - v.second != dist[v.first]) FAIL = true;

        }

    }

}



signed main() {

//    freopen("connect.in", "r", stdin);

//    freopen("connect.out", "w", stdout);

    fastio();



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

    adj.assign(n, vii());



    int l, r, d;

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

        cin >> l >> r >> d; l--, r--;

        adj[l].pb({r, d});

        adj[r].pb({l, -d});

    }



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

        if (!check[i]) {

            dist[i] = 0;

            dfs(i);

        }

    }



    if (!FAIL) cout << "Yes";

    else cout << "No";



//    EXECUTE_TIME();

}