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 il inline
#define RG register
#define ll long long
#define N (500005)

using namespace std;

struct edge{ int nt,to,dis; }g[N];

int head[N],dis[N],vis[N],n,m,num;

il int gi(){
  RG int x=0,q=1; RG char ch=getchar();
  while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();
  if (ch=='-') q=-1,ch=getchar();
  while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q*x;
}

il void insert(RG int from,RG int to,RG int dis){
  g[++num]=(edge){head[from],to,dis},head[from]=num; return;
}

il void dfs(RG int x){
  vis[x]=1;
  for (RG int i=head[x],v;i;i=g[i].nt){
    v=g[i].to;
    if (!vis[v]) dis[v]=dis[x]+g[i].dis,dfs(v);
    else if (dis[v]!=dis[x]+g[i].dis) puts("No"),exit(0);
  }
  return;
}

int main(){
  n=gi(),m=gi();
  for (RG int i=1,l,r,d;i<=m;++i)
    l=gi(),r=gi(),d=gi(),insert(l,r,d),insert(r,l,-d);
  for (RG int i=1;i<=n;++i) if (!vis[i]) dfs(i);
  puts("Yes"); return 0;
}