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: Accepted)

#include<bits/stdc++.h>
using namespace std;
long long d[100005];
int n,m;
bool vis[100005],use[100005];
vector<int>v[100005];
vector<int>e[100005];
void read(int& x)
{
	char c=getchar();
	x=0;
	while(c<'0'||c>'9')c=getchar();
	while(c>='0'&&c<='9')
	{
		x=x*10+c-'0';
		c=getchar();
	}
}
bool dfs(int fa)
{
	vis[fa]=1;
	use[fa]=1;
	for(int i=0;i<v[fa].size();i++)
	{
		int y=v[fa][i];
		int z=e[fa][i];
		if(d[y]>d[fa]+z)
		{
			if(vis[y])return 1;
			d[y]=d[fa]+z;
			if(dfs(y))return 1;
		}
	}
	vis[fa]=0;
	return 0;
}
int main()
{
	read(n);
	read(m);
	for(int i=1;i<=n;i++)
	{
		d[i]=200000000000ll;
	}
	for(int i=1;i<=m;i++)
	{
		int x,y,z;
		read(x);
		read(y);
		read(z);
		v[x].push_back(y);
		e[x].push_back(z);
		v[y].push_back(x);
		e[y].push_back(-z);
	}
	bool ok=0;
	for(int i=1;i<=n;i++)
	{
		if(!use[i])
		{
			d[i]=0;
			if(dfs(i))
			{
				ok=1;
			}
		}
	}
	if(ok)printf("No\n");
	else printf("Yes\n");
	return 0;
}