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: Wrong Answer)

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF=1e9;
const double eps=1e-9;
const int maxn=1e5+10;
const int maxm=2e5+10;
int fa[maxn],dis[maxn];
inline int read(){
	int x=0,flag=1;
	char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')flag=-1,ch=getchar();
	while(isdigit(ch))x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return x*flag;
}
int find(int x){
	if(fa[x]==x)return x;
	int tmp=fa[x];
	fa[x]=find(fa[x]);
	dis[x]+=dis[tmp];
	return fa[x];
}
int main(){
	int i,j,k,m,n;
	n=read();m=read();
	for(i=1;i<=n;i++)fa[i]=i;
	for(i=1;i<=m;i++){
		int x=read(),y=read(),z=read();
		int u=find(x),v=find(y);
		if(u==v){
			if(dis[y]-dis[x]!=z){
				puts("No");
				return 0;
			}
		}
		else{
			dis[v]=dis[y]-dis[x]-z;
			if(dis[v]>0)swap(u,v);
			dis[v]=abs(dis[v]);fa[v]=u;
		}
	}
	puts("Yes");
	return 0;
}