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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <deque>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll,ll> Pll;

#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) (int)x.size()
#define rep(i, n) for(int i=0;i<n;i++)
#define repp(i, n) for(int i=1;i<=n;i++)
#define all(x) x.begin(), x.end()

#define geti1(X) scanf("%d",&X)
#define geti2(X,Y) scanf("%d%d",&X,&Y)
#define geti3(X,Y,Z) scanf("%d%d%d",&X,&Y,&Z)
#define geti4(X,Y,Z,W) scanf("%d%d%d%d",&X,&Y,&Z,&W)

#define GET_MACRO(_1,_2,_3,_4,NAME,...) NAME
#define geti(...) GET_MACRO(__VA_ARGS__, geti4, geti3, geti2, geti1) (__VA_ARGS__)

#define INF 987654321
#define IINF 987654321987654321
#define MAXV 3000500


#define MOD 1234567

int N,M,T,K;

int fix[100500];
vector<Pi> E[100500];
int deg[100500];
int main(){
	geti(N,M);
	repp(i,M){
		int a,b,c; geti(a,b,c);
		E[a].push_back({b,c});
		deg[b]++;
	}

	queue<int> q;
	repp(i,N){
		fix[i] = -1;
		if( deg[i] == 0 ) q.push(i), fix[i] = 0;
		
	}

	while( !q.empty() ){
		int cur = q.front();
		q.pop();
		for(auto e : E[cur]){
			if( fix[e.Fi] == -1 ) {
				fix[e.Fi] = fix[cur] + e.Se;
				q.push(e.Fi);
			}
			else{
				if( fix[e.Fi] != fix[cur] + e.Se ){
					printf("No\n");
					return 0;
				}
			}
		}
	}

	repp(i,N){
		for(auto e : E[i]){
			if( fix[e.Fi] != fix[i] + e.Se ){
				printf("No\n");
				return 0;
			}
		}
	}


	printf("Yes\n");

}