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 <iostream>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<string>
#include<algorithm>
#include<sstream>
#include<cmath>
#include<stack>
#include<map>
#include<cstdio>
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define mp make_pair
#define pb push_back
#define P pair<int,int>
#define ll long long
int n,m;
int l[300000],r[300000],d[300000];
map<int,ll>t;
const ll M=500000000000;

int main(){
cin>>n>>m;
rep(i,n+1)t[i]=-M;
rep(i,m){
cin>>l[i]>>r[i]>>d[i];
l[i]--;
r[i]--;
}

bool ok=1;

rep(i,m){
	if(t[l[i]]==-M&&t[r[i]]==-M){
		t[l[i]]=M;
		t[r[i]]=d[i]+t[l[i]];
	}
	else if(t[l[i]]==-M){
		t[l[i]]=t[r[i]]-d[i];
	}
	else if(t[r[i]]==-M){
		t[r[i]]=t[l[i]]+d[i];
	}
	else{//ここでok=0になりうる
	//	cout<<t[r[i]]<<" "<<t[r[i]]<<" "<<r[i]<<" "<<l[i]<<endl;
		if(d[i]!=t[r[i]]-t[l[i]])ok=0;
	}
}
if(ok)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
	return 0;
}