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)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <numeric>
#include <queue>
#include <map> 
#include <set>
#include <string>
#include <functional>
#include <list>
#include <random>
#include <time.h>
#include <iomanip>
#include <assert.h>
#define int long long
#define double long double
#define oku7 1000000007
#define MAXN (int)1e+5 * 2+1
#define LL_MAX 9223372036854775807	//ない環境用
#define LL_HALFMAX 9223372036854775807 / 2	//ない環境用

using namespace std;
std::mt19937 mt((int)time(0));

int dx[4] = { 0, 1, 0, -1 }; // x軸方向への変位
int dy[4] = { 1, 0, -1, 0 }; // y軸方向への変位

int L[200001], R[200001], D[200001];

int ma[100002];

signed main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < 100002; i++) {
		ma[i] = LL_HALFMAX;
	}
	for (int i = 0; i < M; i++) {
		cin >> L[i] >> R[i] >> D[i];
	}

	int cnt = 0;

	for (int i = 0; i < M; i++) {
		//LもRもまだ位置が決まってなかったら
		if (ma[L[i]] == LL_HALFMAX && ma[R[i]] == LL_HALFMAX) {
			ma[L[i]] = 0;
			ma[R[i]] = D[i];
		}
		//Lが決まってたら
		else if (ma[L[i]] != LL_HALFMAX && ma[R[i]] == LL_HALFMAX) {
			ma[R[i]] = ma[L[i]] + D[i];
		}
		//Rが決まってたら
		else if (ma[R[i]] != LL_HALFMAX && ma[L[i]] == LL_HALFMAX) {
			ma[L[i]] = ma[R[i]] - D[i];
		}
		//両方決まってたら
		else {
			if (ma[R[i]] - ma[L[i]] != D[i])cnt++;
		}
	}
	//はみ出し検査
	int mini = LL_MAX, maxi = -LL_MAX;
	for (int i = 0; i < 100002; i++) {
		if (ma[i] == LL_HALFMAX) continue;
		if (ma[i] < mini) mini = ma[i];
		if (ma[i] > maxi) maxi = ma[i];
	}
	if (maxi - mini > 1000000000) cnt++;

	if (cnt == 0) cout << "Yes\n";
	else cout << "No\n";

	return 0;
}