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 <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
#define REP(i, n) for(int i=0; i<n; i++)
#define MAX_V 100100
#define CONST -1
 
vector <pair<int, int> > L[MAX_V];
queue <pair<int, int> > q;
int V[MAX_V];
bool b[MAX_V];

bool dfs(int v){
    b[v] = true;
    bool ans = true;
    for(int next = 0; next < L[v].size(); next++){
        int nv = L[v][next].first;
        int we = L[v][next].second;
        if(V[nv] == CONST) V[nv] = V[v] + we;
        if(V[nv] == V[v] + we){
            if(!b[nv]) ans = dfs(nv);
        }
        else return false;

        if(!ans) return false;
    }
    return true;
}

int main(){
    int N,M;
    cin >> N >> M;
    int l,r,d;
    REP(i,M){
        cin >> l >> r >> d;
        l--;
        r--;
        L[l].push_back(make_pair(r,d));
        L[r].push_back(make_pair(l,-d));
    }

    bool ans = true;
    fill(V, V+MAX_V, CONST);
    REP(i,N){
        if(!b[i]) {
            V[i] = 0;
            ans = dfs(i);
            if(!ans) break;
        }
    }
    

    if(ans)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}