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)

#include <bits/stdc++.h>
using namespace std;
#define R freopen("555input.txt","r",stdin);
#define N 1000005
typedef long long LL;
int i,j,n,m,st[N],f[N],vis[N],a,b;
LL dis[N];
struct edge
{
    int to;
LL cost;
}t;
vector<edge>v[N];
queue<int>q;
int findSet(int x)
{
    if (x == st[x])
        return x;
    else
        return st[x] = findSet(st[x]);
}
void unionSet(int x, int y)
{
    int fx = findSet(x);
    int fy = findSet(y);
    if (fx != fy)
        st[fx] = fy;
}

int main ()
{
    
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        cin>>a>>b>>t.cost;
        unionSet(a,b);
        if(!f[findSet(a)])
            {f[findSet(a)]=1;q.push(a);vis[a]=1;dis[a]=0;}
        t.to=b;
        v[a].push_back(t);
        t.to=a;
        t.cost=-t.cost;
        v[b].push_back(t);
    }
    while(!q.empty())
    {
        int cur=q.front();
        q.pop();
        int sz=v[cur].size();
        for(i=0;i<sz;i++)
        {
            if(!vis[v[cur][i].to])
            {
                vis[v[cur][i].to]=1;
                q.push(v[cur][i].to);
                dis[v[cur][i].to]=dis[cur]+v[cur][i].cost;
            }
            else if(dis[v[cur][i].to]!=dis[cur]+v[cur][i].cost)
            {
                printf("No");
                return 0;
            }

        }
    }
    printf("Yes");
    return 0;
}