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   M             100005
#define   ll            long long
#define   PI            atan(1.0)*4
#define   mst(x,v)      memset(x,v,sizeof(x))
#define   inf           1e9
#define   rush()        int T;scanf("%d",&T);while(T--)

typedef   pair<int,int> P;

bool f,vis[M];
vector<P> E[M];
int dis[M],n,m;


void dfs(int u,int pre)
{
    if(!f) return;
     int sz=E[u].size();
     pair<int,int>tmp;
     for(int i=0;i<sz;++i)
     {
        tmp=E[u][i];
        if(vis[tmp.first])
        {
            if(dis[u]+tmp.second!=dis[tmp.first])
            {
                f=0;return;
            }
        }
        else
        {
            vis[tmp.first]=1;
            dis[tmp.first]=dis[u]+tmp.second;
            dfs(tmp.first,u);
        }
     }
}

int main() {
	scanf("%d %d",&n,&m);
	
	for(int i=1;i<=m;i++) {
		int l,r,d;
		scanf("%d %d %d",&l,&r,&d);
		E[r].push_back(make_pair(l,d));
		E[l].push_back(make_pair(r,-d));
	}
	
	f=1;
	
	for(int i=1;i<=n;i++) {
		if(!vis[i]) {
			vis[i]=1;
			dfs(i,-1);
		}
	}
	
	puts(f?"Yes":"No");
}