Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,l,r) for(long long i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define REPS(i,n) FOR(i,1,n+1)
#define RFOR(i,l,r) for(long long i=(l);i>=(r);--i)
#define RREP(i,n) RFOR(i,N-1,0)
#define RREPS(i,n) RFOR(i,N,1)
#define int long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
const int INF=1e18;
const int MOD=1e9+7;
const int N_MAX=1e5;
struct Edge{
    int to,cost;
    Edge(int to,int cost):to(to),cost(cost){}
};
int N,M;
vector<Edge>Graph[N_MAX];
bool isvisit[N_MAX];
int d[N_MAX];

bool bad(int k,int dist){
    if(isvisit[k]){
        if(d[k]!=dist)return true;
        return false;
    }
    isvisit[k]=true;
    d[k]=dist;
    for(auto v:Graph[k]){
        if(bad(v.to,d[k]+v.cost))return true;
    }
    return false;
}

void solve(){
    cin>>N>>M;
    REP(i,M){
        int L,R,D;cin>>L>>R>>D;L--;R--;
        Graph[L].eb(R,D);
        Graph[R].eb(L,-D);
    }
    REP(i,N){
        if(isvisit[i])continue;
        if(bad(i,0)){
            cout<<"No"<<endl;
            return;
        }
    }
    cout<<"Yes"<<endl;
}

signed main(){
    solve();
}