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)

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=LLONG_MAX;
const int inf=INT_MAX;
const int nmax=2e5+5;
const int mod=1e9+7;
using namespace std;
int n,m,i,x,y,z,ok,viz[nmax],nr[nmax],sum[nmax];
vector<pair<int,int> >v[nmax];
map<int,int>mp;
void dfs(int x,int p)
{
	if(!ok)return;
	viz[x]=1;
	if(p==mp[x])
    {
   		ok=0;
		return;
	}
	mp[x]=p;
	for(int i=0;i<v[x].size();i++)dfs(v[x][i].fr,x);
}
void dfs1(int x,int s)
{
	if((sum[x]!=s && sum[x]) || !ok)
	{
		ok=0;
		return;
	}
	sum[x]=s;
	for(int i=0;i<v[x].size();i++)dfs1(v[x][i].fr,s+v[x][i].sc);
}
int main()
{
    //freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n>>m;
	for(i=1;i<=m;i++)
	{
		cin>>x>>y>>z;
		v[x].pb({y,z});
		nr[y]++;
	}
	for(i=1;i<=n;i++)
	 if(!viz[i] && v[i].size())
	 {
		ok=1,dfs(i,-1);
		mp.clear();
		if(!ok)rc("No");
	 }
	for(i=1;i<=n;i++)
	 if(viz[i] && !nr[i])
	 {
	 	ok=1,dfs1(i,0);
	 	if(!ok)rc("No");
	 }
	rc("Yes");
	return 0;
}