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 <string>    
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
ll n, m;

struct tosort
{
    ll key;
    ll value[3];

    tosort(ll in_key, ll in_value[3])
    {
        key = in_key;
    	for(ll i = 0; i < 3; i++){
    		value[i] = in_value[i];
    	}
    }
    bool operator<(const tosort &another) const
    {
        return key < another.key;
    };
};

int main() {
	cin >> n >> m;
	ll dist[n];
	for(ll i = 0; i < n; i++){
		dist[i] = 0;
	}
	ll flag = 0;
	if(m != 0){
		vector<tosort> mes;
		ll buf[3];
		for(ll i = 0; i < n; i++){
			cin >> buf[0] >> buf[1] >> buf[2];
			if(buf[0] > buf[1]){
				ll tmp = buf[0];
				buf[0] = buf[1];
				buf[1] = tmp;
			}
			mes.push_back(tosort(buf[0], buf));
		}
		//sort
		sort(mes.begin(), mes.end());
		
		for(ll i = 0; i < n; i++){
			ll l = mes[i].value[0];
			ll r = mes[i].value[1];
			ll d = mes[i].value[2];
			if(dist[r-1] == 0){
				dist[r-1] = dist[l-1]+d;
			}
			else{
				if(dist[l-1]+d != dist[r-1]){
					flag = -1;
					break;
				}
			}
		}
		for(ll i = 0; i < n-1; i++){
			if(dist[i] != 0 && dist[i+1] != 0 && dist[i] > dist[i+1]){
				flag = -1;
			}
		}
	}
	if(flag == -1){
		cout << "No";
	}
	else{
		cout << "Yes";
	}
	return 0;
}