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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)

typedef struct edge{ll s,g,c;}E;
 
ll n,m;
E e[400010];
ll vnow[100010];
ll epre[400010];
 
ll d[100010];
ll que[200010];
void bfs(int k){
	ll qcnt=1;
	que[0]=k;
	rep(i,0,qcnt){
		ll v=que[i];
		for(ll t=vnow[v];t!=-1;t=epre[t]){
			if(!d[e[t].g]){
				que[qcnt++]=e[t].g;
				d[e[t].g]=d[v]+e[t].c;
			}else if(d[e[t].g]!=d[v]+e[t].c){
				puts("No");
				exit(0);
			}
		}
	}
}

int main(){
	scanf("%lld%lld",&n,&m);
	rep(i,0,n)vnow[i]=-1;
	
	rep(i,0,m){
		ll a,b,c;
		scanf("%lld%lld%lld",&a,&b,&c);
		a--;b--;
		e[2*i].s=a;
		e[2*i].g=b;
		e[2*i].c=c;
		epre[2*i]=vnow[a];
		vnow[a]=2*i;
		
		e[2*i+1].s=b;
		e[2*i+1].g=a;
		e[2*i+1].c=-c;
		epre[2*i+1]=vnow[b];
		vnow[b]=2*i+1;
	}
	ll cnt=0;
	rep(i,0,n)if(!d[i]){
		d[i]=++cnt<<32;
		bfs(i);
	}
	puts("Yes");
	return 0;
}