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 <queue>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
const int maxn=200010; 
const int mmin=-200000000;
int N,M;
int X[maxn],Li[maxn],Ri[maxn],Di[maxn];
struct message
{
	int l;
	int r;
	int d;
	int circle;
};
message a[maxn];
queue <message> q;
int main()
{
//		freopen("in.txt","r",stdin);
	scanf("%d%d",&N,&M);
		while(!q.empty()) q.pop();
		for(int i=0;i<maxn;i++)
			X[i]=mmin;
		for(int i=0;i<M;i++)
		{
			scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].d);
			a[i].circle=1;
			q.push(a[i]);
//			printf("push l=%d r=%d d=%d  circle=%d\n",a[i].l,a[i].r,a[i].d,a[i].circle);
		}
		int flag=0;
//		for(int i=0;i<M;i++)
//			printf("Xi=%d\n",X[i]);
		int nowcircle=0;
		while(!q.empty())
		{
			message now=q.front();
			q.pop(); 
			int li=now.l,ri=now.r,di=now.d,ci=now.circle;
//均未出现该关系点,初始化
//均未初始化
//均初始化 
			if(ri>N||li>N)
			{
				flag=1;
				break;
			}
			else if(X[ri]!=mmin&&X[li]!=mmin)
			{
//				printf("Xl[%d]=%d   Xr[%d]=%d   X %d\n",li,X[li],ri,X[ri],X[2]);
				int tempXi=X[li]+di;
				if(tempXi!=X[ri])
				{
					flag=1;
//					printf("argue l=%d r=%d d=%d  circle=%d\n",now.l,now.r,now.d,now.circle);
					break;
				}
			}
			else if(X[ri]!=mmin) 
			{
				X[li]=X[ri]-di;
//				printf("set left l=%d r=%d d=%d  circle=%d\n",now.l,now.r,now.d,now.circle);
			}
			else if(X[li]!=mmin) 
			{
				X[ri]=X[li]+di;
//				printf("set right l=%d r=%d d=%d  circle=%d\n",now.l,now.r,now.d,now.circle);
			}
			else if(nowcircle!=ci)
			{
				nowcircle=ci;
				X[li]=0;
				X[ri]=X[li]+di;
//				printf("set zero l=%d r=%d d=%d  circle=%d\n",now.l,now.r,now.d,now.circle);
			}
			else
			{
				message next;
				next.l=li;
				next.r=ri;
				next.d=di;
				next.circle=ci+1;
				q.push(next);
//				printf("roll back l=%d r=%d d=%d  circle=%d\n",next.l,next.r,next.d,next.circle);
			}
		}
		if(flag) printf("No\n");
		else printf("Yes\n");

		
	return 0;
}