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>
#define maxn 100100
using namespace std;
int fa[maxn];
long long a[maxn];
int ff(int x)
{
    vector<long long>t,u;
    while(x!=fa[x])
    {
        t.push_back(x);
        u.push_back(a[x]);
        x=fa[x];
    }
    long long s=0;
    for(int i=(int)t.size()-1;i>=0;i--)
    {
        s+=u[i];
        a[t[i]]=s;
        fa[t[i]]=x;
    }
    return x;
}
int main()
{
    cin.tie(0);ios_base::sync_with_stdio(false);
    int n,m,l,r,d;
    cin>>n>>m;
    for(int i=1;i<=n;i++)fa[i]=i;
    while(m--)
    {
        cin>>l>>r>>d;
        ff(l);
        ff(r);
        if(fa[l]==fa[r])
        {
            if(a[r]-a[l]==d)continue;
            else {cout<<"No"<<endl;exit(0);}
        }
        a[fa[r]]=a[l]-a[r]+d;
        fa[fa[r]]=fa[l];
    }
    cout<<"Yes"<<endl;
return 0;
}