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 <iostream>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <stack>
#include <functional>

using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> P;
typedef vector<P> vpl;
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(ll i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int mod = 1e9+7;
const int MAX = 5100000;
const int V = 100005;
ll dy[8] = {1,0,-1,0,1,-1,1,-1};
ll dx[8] = {0,1,0,-1,1,-1,-1,1};
const double pi = acos(-1);
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){cout << a << " " << b << endl;}
#define INT_MAX 1e9+7
vector<ll> fac(MAX), finv(MAX), inv(MAX);

int main(){
    ll n,m; cin >> n >> m;
    vector<vpl> G(n);
    rep(i,m){
        ll a,b,c; cin >> a >> b >> c; a--; b--;
        G[a].emplace_back(b,c);
        G[b].emplace_back(a,-c);
    }
    set<ll> se;
    rep(i,n) se.insert(i);
    vl x(n,inf);
    while(!se.empty()){
        ll s = *se.begin(); se.erase(s);
        if(x[s] != inf) continue;
        if(G[s].empty()) continue;
        stack<ll> q;
        q.push(s);
        x[s] = 0;
        while(!q.empty()){
            ll u = q.top(); q.pop();
            for(auto v : G[u]){
                if(x[v.first]==inf){
                    x[v.first] = x[u] + v.second;
                    q.push(v.first);
                }
                else{
                    if(x[v.first] != x[u] + v.second){
                        puts("No");
                        return 0;
                    }
                }
            }
        }
    }
    puts("Yes");
}