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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<algorithm>
#include<cmath>
#include<iomanip>
#include<iostream>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
#include<vector>
using uint = unsigned int;
using ll = long long;
enum : int { M = (int)1e9 + 7 };
enum : ll { MLL = (ll)1e18L + 9 };
using namespace std;
#ifdef LOCAL
#include"rprint2.hpp"
#include"debug_deque.hpp"
#define vector DebugDeque
#else
#define FUNC(name) template <ostream& out = cout, class... T> void name(T&&...){ }
FUNC(prints) FUNC(printe) FUNC(printw) FUNC(printew) FUNC(printb) FUNC(printd) FUNC(printde);
#endif
template <class T>
istream& operator >> (istream& in, vector<T>& v){ for(auto& e : v){ in >> e; } return in; }

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    vector<vector<pair<int, int>>> rds(n + 1);
    for(int i = 0; i < m; i++){
        int l, r, d;
        scanf("%d %d %d", &l, &r, &d);
        rds[l].emplace_back(r, d);
        rds[r].emplace_back(l, -d);
    }
    vector<char> checked(n + 1);
    vector<int> poss(n + 1, -1);
    bool ok = true;
    for(int i = 1; i <= n && ok; i++){
        printd(checked);
        if(checked[i]){ continue; }
        queue<int> q;
        checked[i] = true;
        q.push(i);
        while(ok && q.size()){
            int idx = q.front(); q.pop();
            for(auto& rd : rds[idx]){
                int pos = poss[idx] == -1 ? 0 : poss[idx];
                int i2 = rd.first;
                int p2 = pos + rd.second;
                if(!checked[i2]){
                    checked[i2] = true;
                    q.push(i2);
                }
                if(poss[i2] == -1){
                    poss[i2] = p2;
                }else{
                    if(poss[i2] != p2){
                        ok = false;
                    }
                }
            }
            printd(poss);
        }
    }
    cout << (ok ? "Yes" : "No") << '\n';
}