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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

typedef struct WeightedUnionFind{
  int *parent;
  int *distance;
  int n;
} UnionFind;

UnionFind* newUnionFind(int n){
  UnionFind *u=(UnionFind *)calloc(1,sizeof(UnionFind));
  u->parent=(int *)calloc(2*n,sizeof(int));
  u->distance=u->parent+n;
  u->n=n;
  int i;
  for(i=0;i<n;i++){
    u->parent[i]=i;
    u->distance[i]=0;
  }
  return u;
}

int root(UnionFind *u,int x){
  if(u->parent[x]==x) return x;
  int p=u->parent[x];
  u->parent[x]=root(u,u->parent[x]);
  u->distance[x]+=u->distance[p];
  return u->parent[x];
}

int same(UnionFind *u,int x,int y){
  return root(u,x)==root(u,y);
}

int weight(UnionFind *u,int x){
  root(u,x);
  return u->distance[x];
}

int diff(UnionFind *u,int x,int y){
  return weight(u,y)-weight(u,x);
}

void unite(UnionFind *u,int x,int y,int w){
  w+=weight(u,x);
  w-=weight(u,y);
  x=root(u,x);
  y=root(u,y);
  if(x==y) return;
  u->parent[y]=x;
  u->distance[y]+=w;
}

void run(void){
  int n,m;
  scanf("%d%d",&n,&m);
  UnionFind *u=newUnionFind(n+1);
  while(m--){
    int l,r,d;
    scanf("%d%d%d",&l,&r,&d);
    if(!same(u,l,r)){
      unite(u,l,r,d);
    } else if(diff(u,l,r)!=d){
      puts("No");
      return;
    }
  }
  puts("Yes");
}

int main(void){
  run();
  return 0;
}