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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,s,e) for (int i=s;i<(e);i++)
#define All(v) (v).begin(),(v).end()
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
using namespace std;
typedef long long llint;
typedef pair<int, int> P;
const int MOD = (int)1e9 + 7;
const int INF = numeric_limits<int>::max();

struct edge
{
    int to;
    int cost;
    edge(int to, int cost) : to(to), cost(cost) {}
};
typedef vector< vector < edge > > Graph;

void dfs(Graph &G, int st, vector<int> &x){
    x[st] = 0;

    stack<int> S;
    int N = G.size();
    static vector<vector<edge>::iterator> itr(N);
    if(st == 0) REP(i, N){ itr[i] = G[i].begin(); }
    vector<int> done(N, 0);
    S.push(st);
    while(!S.empty()){
        int u = S.top();
        if(itr[u] == G[u].end()){
            S.pop();
            done[u] = 2;
        }else{
            edge v = *itr[u]++;
            if(!done[v.to]){
                done[v.to] = 1;
                S.push(v.to);
                x[v.to] = x[u] + v.cost;
            }
        }
    }
}


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

    int N, M;
    cin >> N >> M;

    Graph G(N);
    REP(i, M){
        int l, r, d;
        cin >> l >> r >> d;
        l--; r--;
        G[l].pb(edge(r, d));
        G[r].pb(edge(l, -d));
    }
    vector<int> x(N, INF);
    REP(i, N){
        if(x[i] == INF) dfs(G, i, x);
    }
    REP(l, N){
        // cout << x[l] << "\n";
        for(auto e : G[l]){
            if(x[e.to] - x[l] != e.cost){
                puts("No");
                return 0;
            }
        }
    }
    puts("Yes");

    return 0;
}