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 <stdio.h>

using namespace std;

const int MAX_N = 100000;
int par[MAX_N + 1];
int rnk[MAX_N + 1];
int dis[MAX_N + 1];

int root(int x)
{
	if(par[x]==x)return x;
	return par[x] = root(par[x]);
}

int find(int x)
{
	if(par[x]==root(x)) return 0;
	return dis[x] = dis[x]+find(par[x]);
}

int merge(int L,int R,int D)
{	
	int x = L,y = R;
	L = root(L);
	R = root(R);
	int L_d = find(x);
	int R_d = find(y);
	if(L==R){
		if(R_d-L_d==D) return 1;
		else return 0;
	}
	if(rnk[L]<rnk[R])
	{
		par[L] = R;
		dis[L] = R_d + D - L_d;
	}
	else {
		par[R] = L;
		dis[R] = L_d + D - R_d;
		if(rnk[L]==rnk[R])rnk[L]++;
	}
	return 1;
}

int main()
{
	int n,m,i;
	scanf("%d %d",&n,&m);
	
	for(i=1;i<=n;i++)par[i]=i;
	
	for(i=0;i<m;i++)
	{
		int L,R,D;
		scanf("%d %d %d",&L,&R,&D);
		if(!merge(L,R,D)){
			printf("No"); return 0;
		}
		
	}
	printf("Yes");
	return 0;
}