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)

//状态:
//题号:
#include <iostream>
#include <stdio.h>
#include <fstream>
#include <algorithm>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <iomanip>
#include <cassert>
#include <complex>
#include <functional>
#include <list>
using namespace std;
typedef pair<int,int> pii;
const int maxn=100005;
int d[maxn];
int n,m;
map<pii,int> dis;
vector<int> nei[maxn];
bool vis[maxn];
void dfs(int pos,int par=-1){
	//cout<<pos<<" "<<par<<endl;
	if(vis[pos]||pos>=n)return;
	d[pos]=(par==-1?0:d[par]+dis[pii(pos,par)]);
	vis[pos]=true;
	for(int i=0;i<nei[pos].size();i++){
		if(!vis[nei[pos][i]])dfs(nei[pos][i],pos);
	}
	dfs(pos+1,pos);
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		a--;
		b--;
		dis[pii(a,b)]=c;
		dis[pii(b,a)]=-c;
		nei[a].push_back(b);
		nei[b].push_back(a);
	}
	memset(d,-1,sizeof(d));
	d[0]=0;
	dfs(0);
	//for(int i=0;i<n;i++)cout<<d[i]<<" ";
	//cout<<endl;
	for(map<pii,int>::iterator it=dis.begin();it!=dis.end();it++){
		if(it->second!=-1&&it->second!=d[it->first.first]-d[it->first.second]){
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	return 0;
}