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 <bits/stdc++.h>
 
int st, x, l, r, d, sw, n, m, dd, tmp;
 
struct S{
	int d;
	int std;
}s[100010];
 
int stds[100010];

int root(int i){
	if(s[i].std == i) return i;
	return root(s[i].std);
}
 
int main(){
	scanf("%d %d", &n, &m);
	
	sw = 1;
	
	for(int i=1; i<=m; i++){
		scanf("%d %d %d", &l, &r, &d);
		
		if(s[l].std==0 && s[r].std==0){
			s[l].std = s[r].std = l;
			stds[l] += 2;
			s[r].d = d;
		}
		else if(s[l].std == 0){
			s[l].std = s[r].std;
			s[l].d = s[r].d - d;
			stds[s[r].std]++;
		}
		else if(s[r].std == 0){
			s[r].std = s[l].std;
			s[r].d = s[l].d + d;
			stds[s[l].std]++;
		}
		else {
			if(s[l].std != root(l)){
				s[l].d += s[s[l].std].d;
				s[l].std = root(l);
			} 
			if(s[r].std != root(r)){
				s[r].d += s[s[r].std].d;
				s[r].std = root(r);
			}
			if(s[r].std != s[l].std){
				if(stds[s[r].std] < stds[s[l].std]) x = r; // 적은게 x 
				else x = l;
				dd = s[x].d;
				st = s[x].std;
				stds[s[r+l-x].std] += stds[s[x].std];
				tmp = s[r+l-x].d - dd + (x==r? d:-d);
			}
			if(s[r].d != s[l].d + d) {
				sw = 0;
				break;
			}
		}
	}
	if(sw) printf("Yes");
	else printf("No");
	
	return 0;
}