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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define LL long long
using namespace std;
struct A{
	int into,v;
	A(){}
	A(int into,int v):into(into),v(v){}
};
const int maxn=100011;
const LL maxk=100000000000LL;
vector<A>a[maxn];
LL c[maxn];
int n,m;
bool flag;
bool sf[maxn];
bool diansf[maxn];
bool spfa(int x){
	memset(c,63,sizeof(c));
	memset(sf,false,sizeof(sf));
	//cout<<c[0]<<endl;
	c[x]=0;
	sf[x]=true;
	queue<int>q;
	q.push(x);
	while(!q.empty()){
		int xx=q.front();
		q.pop();
		diansf[xx]=true;
		sf[xx]=false;
		for(int i=0;i<a[xx].size();i++){
			if(c[a[xx][i].into]!=a[xx][i].v+c[xx]){
				if(c[a[xx][i].into]<=maxk)return false;
				c[a[xx][i].into]=a[xx][i].v+c[xx];
				if(!sf[a[xx][i].into]){
					q.push(a[xx][i].into);
					sf[a[xx][i].into]=true;
				}
			}
		}
	}
	return true;
}
int main(int argc, char *argv[])
{
	int x,y,z;
	flag=true;
	memset(diansf,false,sizeof(diansf));
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d%d",&x,&y,&z);
		a[x].push_back(A(y,z));
	}
	for(int i=1;i<=n;i++){
		if(!diansf[i]){
			flag=spfa(i);
			if(!flag)break;
		}
	}
	if(flag)puts("Yes");
	else puts("No");
}