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)

//ヘッダー
#include<bits/stdc++.h>
using namespace std;

//型定義
typedef long long ll;

//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1e+9+7 //10^9+7
#define MAXR 100000 //10^5:配列の最大のrange(素数列挙などで使用)

//REPマクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define REP2(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define REPD2(i,a,b) for(ll i=a;i>=(ll)(b);i--)

//vectorの扱い
#define ALL(x) (x).begin(),(x).end() //sortなどの引数省略
#define SIZE(x) ((ll)(x).size()) //size
#define MAX(x) *max_element(ALL(x)) //最大値
#define MIN(x) *min_element(ALL(x)) //最小値

//グローバル変数


//ライブラリ挿入



int main(){
    int N,M;
    cin >> N >> M;
    vector<vector<pair<int,ll>>> dist(N);
    vector<ll> x(N,INF);
    ll l,r,d;
    x[0]=0;
    REP(i,M){
        cin >> l >> r >> d;
        l--;
        r--;
        if(l<r){
            dist[l].push_back({r,d});
        }else{
            dist[r].push_back({l,-d});
        }
    }

    int i_next;
    ll x_next;
    bool flag=true;
    ll x_min=x[0];
    ll x_max=x[0];
    REP(i,N){
        REP(j,dist[i].size()){
            i_next=dist[i][j].first;
            x_next=x[i]+dist[i][j].second;
            if(x[i_next]!=x_next&&x[i_next]!=INF){
                flag=false;
            }else{
                x[i_next]=x_next;
                x_max=max(x_max,x_next);
                x_min=min(x_min,x_next);
            }
            if(!flag) break;
        }
        if(!flag) break;
    }

    if(x_max-x_min>1e+9) flag=false;
    if(flag){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
}