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<cstdio>
#include<iostream>
#include<queue>
#define ll long long
using namespace std;
int n,m,doe;
int pre[804040],son[804040],now[201010];
ll f[201010],v[804040];
bool br[201010],ok[201010];
int du[201010];
int que[20101010];
void add(int x,int y,ll z)
{
	++doe;
	pre[doe]=now[x];
	now[x]=doe;
	son[doe]=y;
	v[doe]=z;
}
bool spfa(int st)
{
	int head=1,tail=1;
	que[1]=st;
	f[st]=0;
	ok[st]=1;
	br[st]=1;
	while (head<=tail)
	{
		int x=que[head];
		br[x]=0;
		++du[x];
		//printf("%d %d\n",x,du[x]);
		if (du[x]>=n/2) return 1;
 
		int p=now[x];
		while (p)
		{
			int y=son[p];
			if (f[y]>f[x]+v[p] || ok[y]==0)
			{
				ok[y]=1;
				f[y]=f[x]+v[p];
				if (br[y]==0)
				{
					br[y]=1;
					++tail;
					que[tail]=y;
				}
			}
			p=pre[p];
		}
		++head;
	}
	return 0;
}
int main()
{
//	freopen("D.in","r",stdin);
	scanf("%d%d",&n,&m);
	int x,y,z;
	for (int i=1;i<=m;++i)
	{
		scanf("%d%d%d",&x,&y,&z);
		add(x,y,z);
		add(y,x,-z);
	}
//	printf("a\n");
	for (int i=1;i<=n;++i)
		if (ok[i]==0)
			if (spfa(i))
			{
				printf("No\n");
				return 0;
			}
	printf("Yes\n");
	return 0;
}