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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int INF = 1e9;
const int MOD = 1000000007;
const int MAX = 1000000;

int main(){
        int n,m;
        cin >> n >> m;
        vector<ll> dis(n,0);
        vector<bool> seen(n,false);
        bool ok = true;
        rep(i,m){
                int l,r;
                ll d;
                cin >> l >> r >> d;
                l--;r--;
                if(i==0){
                        dis[l] = 0LL;
                        dis[r] = d;
                }else{
                        if(seen[l] && seen[r]){
                                if(dis[r] - dis[l] != d){
                                        ok = false;
                                        break;
                                }
                        }else if(seen[l]){
                                dis[r] = dis[l] + d;
                        }else if(seen[r]){
                                dis[l] = dis[r] - d;
                        }
                }
                seen[l] = seen[r] = true;
        }

        //rep(i,n) cout << dis[i] << " " ;
        if(!ok){
                cout << "No" << endl;
                return 0;
        }


        ll mx=0,mn=0;
        rep(i,n){
                mx = max(mx,dis[i]);
                mn = min(mn,dis[i]);
        }
        if(mx-mn > 1000000000){
                cout << "No"  << endl;
                return 0;
        }

        cout << "Yes" << endl;
        return 0;
}