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<stdio.h>
#define INF 100100100
int main(void){
	int n,m;
	int l,r,d;
	signed x[100100]={0};
	int list[500100]={0};
	int node[500100];
	int stack[500100]={0};
	int slast=0;
	signed cost[500100];
	int i;
	int llast=0;
	scanf("%d%d",&n,&m);
	llast=n+1;
	for(i=0;i<m;i++){
		scanf("%d%d%d",&l,&r,&d);
		list[llast]=list[l];
		list[l]=llast;
		node[llast]=r;
		cost[llast]=d;
		llast++;
		list[llast]=list[r];
		list[r]=llast;
		node[llast]=l;
		cost[llast]=-d;
		llast++;
	}
	for(i=0;i<=n;i++){
		x[i]=INF;
	}
	int flg=0,look,p;
	i=1;
	while(i<=n){
		x[i]=1;
		stack[0]=i;
		slast=1;
		while(slast){
			slast--;
			look=stack[slast];
			p=list[look];
			while(p){
				if(x[node[p]]==INF){
					x[node[p]]=x[look]+cost[p];
					stack[slast]=node[p];
					slast++;
				}else{
					if(x[node[p]]!=x[look]+cost[p]) flg=1;
				}
				p=list[p];
			}
		}
		while(i<=n&&x[i]!=INF) i++;
	}
	if(flg){
		puts("No");
	}else{
		puts("Yes");
	}
	return 0;
}