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)

#define debug_interval ','
#define dump_interval ' '
#define debug_toggle 1
//{
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define INT_MAX LLONG_MAX
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define fi first
#define se second
#define mp make_pair
#define rev reverse
#define dans dump(ans)
 
//{
inline int toInt(string s){int v;istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x){ostringstream sout;sout<<x;return sout.str();}
template<class...A> inline void dump(){cout<<endl;}
template<class...A> inline void dump_rest() {cout<<endl;}
template<class T, class...A> inline void dump_rest(const T& first, const A&... rest){cout<<dump_interval<<first;dump_rest(rest...);}
template<class T,class...A> inline void dump(const T&first,const A&...rest){cout<<first;dump_rest(rest...);}
template<class...A> inline void debug(){cerr<<endl;}
template<class...A> inline void debug_rest() {cerr<<endl;}
template<class T, class...A> inline void debug_rest(const T& first, const A&... rest){cerr<<debug_interval<<first;debug_rest(rest...);}
template<class T,class...A> inline void debug(const T&first,const A&...rest){if(debug_toggle)cerr<<first,debug_rest(rest...);}
//}

typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}
template<int m>class mint{private:int i;public:mint() : i(0){}mint(int i): i((i%m+m)%m){}mint operator+(const mint& o){return o.i+i;}mint operator*(const mint& o){return o.i*i;}mint operator-(){return -i;}operator int() {return i;}};

//}

unordered_map<int,int> ans;
int n,m,l,r,d;

main(){
	cin>>n>>m;
	cin>>l>>r>>d;
	ans.insert(mp(l,0));
	ans.insert(mp(r,ans[l]+d));
	rep(i,n-1){
		cin>>l>>r>>d;
		if(ans.count(r)==0)ans.insert(mp(r,ans[l]+d));
		else{
			if(ans[r]!=ans[l]+d){
				cout<<"No"<<endl;
				return 0;
			}
		}
	}
	cout<<"Yes"<<endl;
}