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: Accepted)

#include<algorithm>
#include<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

template < typename T >
void vprint(T &V){
	for(auto v : V){
    	cout << v << " ";
	}
	cout << endl;
}

const ll mod = 1e9 + 7;
const ll inf = 1e18;

struct Edge{
    ll to;
    ll cost;
    Edge(ll to, ll cost): to(to), cost(cost) {}
    Edge(){}
};

const int N_MAX = 100010;
vector<vector<Edge> > G;
bool visited[N_MAX];
ll x[N_MAX] = {};
bool is_consistent = true;

void print_x(){
    FOR(i, 0, 10){
        cout << x[i] << " ";
    }
    cout << endl;
}

// 矛盾チェック
void dfs(ll current, ll prev, ll position){
    
    visited[current] = true;
    x[current] = position;

    // cout << "dfs " << current << " " << prev << endl;
    // print_x();

    for(auto edge : G[current]){
        ll next_x = x[current] + edge.cost;
        if(visited[edge.to]){
            // 距離の整合性チェック
            if(x[edge.to]!=next_x){
                is_consistent = false;
            }
            continue;
        }else{
            dfs(edge.to, current, next_x);
        }
    }
    return;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll N, M;
    cin >> N >> M;

    FOR(i, 0, N_MAX){
        visited[i] = false;
        x[i] = 0;
    }

    G.resize(N+1);

    FOR(i, 0, M){
        ll l, r, d;
        cin >> l >> r >> d;
        
        G[l].push_back(Edge(r, d));
        G[r].push_back(Edge(l, -d));
    }

    FOR(i, 1, N+1){
        if(!visited[i]){
            dfs(i, -1, 0);
        }
    }

    if(is_consistent){
        p_yes();
    }else{
        p_no();
    }
    
    return 0;
}