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<iostream>
#include<cstdio>
using namespace std;
struct Edge
{
	int to,next,cost;	
}e[400005];
int head[100005],cnt=0;
bool vis[100005];
int d[100005];
bool use[100005];
void addedge(int from,int to,int cost)
{
	e[++cnt].to=to;
	e[cnt].cost=cost;
	e[cnt].next=head[from];
	head[from]=cnt;
}
bool flag=0;
void dfs(int x)
{
	vis[x]=1;
	for(int i=head[x];i;i=e[i].next)
	{
		int w=e[i].cost,v=e[i].to;
		if(d[v]>d[x]+w)
		{
			d[v]=d[x]+w;
			if(vis[v])
			{
				flag=1;
				return;
			}
			dfs(v);	
		}	
	} 
	vis[x]=0;
	return;
}
inline int read()
{
	int s=0,f=1;char x=getchar();
	while(!isdigit(x))
	{
		if(x=='-') f=-1;
		x=getchar();
	}
	while(isdigit(x))
	{
		s=(s<<1)+(s<<3)+(x^48);
		x=getchar();
	}
	return s*f;
}
int main()
{
	int n,m;
	n=read();m=read();
	for(int i=1;i<=m;i++)
	{
		int x,y,z;
		x=read();y=read();z=read();
		addedge(y,x,z);
		addedge(x,y,-z);
	}
	flag=0;
	for(int i=1;i<=n;i++)
	{
		d[i]=0;
		if(!use[i])
		{
			use[i]=1;
			dfs(i);
		}	
	}
	if(flag) cout<<"No";
	else cout<<"Yes";
	return 0;
}