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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define gc getchar
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define x1 uvuvwevwevwe
#define y1 onyetenyevwe
#define x2 ugwemubwem
#define y2 ossas
typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
template <typename T> void scan(T &angka){
	angka=0;char input=gc();T kali=1;
	while(!(48<=input&&input<=57)){	if(input=='-')	kali=-1;input=gc();}
	while(48<=input&&input<=57)	angka=(angka<<3)+(angka<<1)+input-48,input=gc();angka*=kali;
}
template <typename T1,typename T2> void scan(T1 &angka1,T2 &angka2){
	scan(angka1);scan(angka2);
}
const int MAXN=100000;
int pos[MAXN+5],n,m;
//sudah punya posisi atau tidak
bool sudah[MAXN+5],visited[MAXN+5];
vector <pii> node[MAXN+5];
bool solve=true;
void dovisit(int now){
	if(visited[now])
		return;
	visited[now]=true;
	if(!sudah[now])
		pos[now]=0;
	sudah[now]=true;
	for(auto v:node[now])
	{
		if(!sudah[v.fi])
		{
			pos[v.fi]=pos[now]+v.se;
			sudah[v.fi]=true;
		}
		else
			solve=solve&&(pos[v.fi]==pos[now]+v.se);
		dovisit(v.fi);
	}
}
int main()
{
	memset(sudah,false,sizeof(sudah));
	scan(n,m);
	for(int i=0;i<m;i++)
	{
		int l,r,jarak;
		scan(l,r),scan(jarak);
		node[l].pb(mp(r,jarak));
		node[r].pb(mp(l,-jarak));
	}
	for(int i=1;i<=n;i++)
		dovisit(i);
	if(!solve)
		cout<<"No"<<endl;
	else
		cout<<"Yes"<<endl;
}