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<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;
int hd[100010],nxt[400010],edg[400010],ver[400010];
int n,m;
int l,r,dd;
int tot,nuuum;
int dist[100010],visnum[100010];
bool ins[100010];
int mmnb;
inline int rd()
{
    char ch;
    int res;
    while((ch=getchar())>'9'||ch<'0');
    res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=(res<<1)+(res<<3)+ch-'0';
    return res;
}
inline void add_edge(int x,int y,int z)
{
    ver[++tot]=y,edg[tot]=z,nxt[tot]=hd[x],hd[x]=tot;
}
int main()
{
    memset(dist,0x3f,sizeof(dist));
    n=rd(),m=rd();
    if(m==0) {printf("Yes");return 0;}
    for(int i=1;i<=m;i++)
    {
        l=rd(),r=rd(),dd=rd();
        add_edge(r,l,dd);
        add_edge(l,r,-dd);
    }
    for(int ttt=1;ttt<=n;ttt++)
    {
        if(dist[ttt]==1061109567)
        {
            ++nuuum;
            int s=ttt;
            queue<int> que;
    int summ=1;
    que.push(s);
    visnum[s]=nuuum;
    while(!que.empty())
    {
        int ttemp=que.front();
        que.pop();
        for(int m=hd[ttemp];m;m=nxt[m])
        {
            int y=ver[m];
            if(visnum[y]==0) {visnum[y]=nuuum,++summ;que.push(y);}
        }
    }
    queue<int> q;
    q.push(s);
    dist[s]=0;
    while(!q.empty())
    {
        int y=q.front();
        q.pop();
        ins[y]=0;
        for(int i=hd[y];i;i=nxt[i])
        {
            int z=ver[i];
            if(visnum[z]==nuuum)
            {
                if(dist[z]>dist[y]+edg[i])
                {
                	if(dist[z]==1061109567) {dist[z]=dist[y]+edg[i];q.push(z);}
                }
                else
                if(dist[z]!=dist[y]+edg[i]) {printf("No");return 0;}
            }
        }
    }
        }
    }
    printf("Yes");
}