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: Wrong Answer)

#include <cstdio>
#include <queue>
const int N=1e5+5;
const int M=2e5+5;
const long long inf=1LL<<60;
int n,m,tot,hd[N],ter[M],nxt[M],val[M];
long long dis[N];
bool vis[N],flg[N];
std::queue<int> q;

void add(int u,int v,int w) {
	ter[++tot]=v;
	val[tot]=w;
	nxt[tot]=hd[u];
	hd[u]=tot;
}
bool spfa(int s) {
	q.push(s),vis[s]=1,dis[s]=0;
	while(!q.empty()) {
		int u=q.front(); q.pop();
		for(int i=hd[u];i;i=nxt[i]) {
			int v=ter[i];
			if(dis[v]==inf) {
				dis[v]=dis[u]+val[i];
				flg[v]=1;
				if(!vis[v]) q.push(v);
			} else {
				if(dis[v]!=dis[u]+val[i]) return 0;
			}
		}
		vis[u]=0;
	}
	return 1;
}
int main() {
	scanf("%d%d",&n,&m);
	for(int u,v,w;m;--m) {
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w),add(v,u,-w);
	}
	for(int i=1;i<=n;++i) dis[i]=inf;
	for(int i=1;i<=n;++i) {
		if(!flg[i]) {
			flg[i]=1;
			if(!spfa(i)) return puts("No"),0;
		}
	}
	puts("Yes");
	return 0;
}