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)

#include <iostream>
#include <fstream>
#include <cmath>  
#include <cstdlib>
#include <ctime>
#include <algorithm>
#include <numeric>
#include <functional>
#include <string> 
#include <vector>
#include <bitset>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;

#define REP(i,n) for(int i = 0; i < int(n); i++)
#define REREP(i,n) for(int i = int(n - 1); i >= 0; i--)
#define FOR(i, m, n) for(int i = int(m);i < int(n);i++)
#define REFOR(i, m, n) for(int i = int(n - 1);i >= int(m);i--)
#define ALL(obj) (obj).begin(),(obj).end()
#define fi first 
#define se second

#define VI vector<int>
#define VVI vector<vector<int>>
#define VVVI vector<vector<vector<int>>>
#define VD vector<double>
#define VVD vector<vector<double>>
#define VVVD vector<vector<vector<double>>>
#define VL vector<ll>
#define VVL vector<vector<ll>>
#define VVVL vector<vector<vector<ll>>>
#define VB vector<bool>
#define VVB vector<vector<bool>>
#define VVVB vector<vector<vector<bool>>>
#define VS vector<string>
#define VVS vector<vector<string>>
#define VVVS vector<vector<vector<string>>>

#define PII pair<int,int>
#define PDD pair<double,double>
#define PLL pair<ll,ll>
#define VPII vector<pair<int,int>>
#define VVPII vector<vector<pair<int,int>>>
#define VVVPII vector<vector<vector<pair<int,int>>>>
#define VPLL vector<pair<ll,ll>>
#define VVPLL vector<vector<pair<ll,ll>>>
#define VVVPLL vector<vector<vector<pair<ll,ll>>>>

const ll MOD = (ll)1e9 + 7;
const ll HINF = (ll)1e18;
const ll LINF = (ll)1e9;
const long double PI = 3.1415926535897932384626433;


void YN(bool flag) {
	cout << ((flag) ? "YES" : "NO") << endl;
}

void Yn(bool flag) {
	cout << ((flag) ? "Yes" : "No") << endl;
}

void yn(bool flag) {
	cout << ((flag) ? "yes" : "no") << endl;
}

//union-find
template<class T> class Union_Find {
public:
	vector<int> prnt, rank;
	vector<T> diff;
	
	Union_Find(int N) : prnt(vector<int>(N)), rank(vector<int>(N,0)), diff(vector<T>(N,0)) {
		for (int i = 0; i < N; ++i) prnt[i] = i;
	}
	
	int root(int x) {
		if (prnt[x] == x) return x;
		else {
			diff[x] += diff[prnt[x]];
			return prnt[x] = root(prnt[x]);
		}
	}
	
	T weight(int x) {
		root(x);
		return diff[x];
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}
	
	void unite(int x, int y, T d) {
		d += weight(x) - weight(y);
		x = root(x);
		y = root(y);
		if (x == y) return;
		if (rank[x] < rank[y]) swap(x, y), d *= -1;
		if (rank[x] == rank[y]) ++rank[x];
		prnt[y] = x;
		diff[y] = d;
	}
};

int main() {
	//ll N, M, X; cin >> N >> M >> X;
	//vector<pair<ll, pair<ll, ll>>> edge(M);
	//REP(i, M) {
	//	ll U, V, W;
	//	cin >> U >> V >> W;
	//	U--; V--;
	//	edge[i] = { W, {U,V} };
	//}

	int N, M; cin >> N >> M;
	Union_Find<int> UF(N);
	bool flg = true;
	REP(i, M) {
		int L, R, D;
		cin >> L >> R >> D;
		L--; R--;
		if (UF.same(L, R)) {
			if (D != abs(UF.weight(L) - UF.weight(R))) flg = false;
		}
		else UF.unite(L,R,D);
	}
	Yn(flg);
	
	return 0;
}