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: Runtime Error)

/**
*
* Author: MARS
* Lang: GNU C++14
*
**/

#include<bits/stdc++.h>
using namespace std;

#define F first
#define S second
#define ii pair < int , int >
#define LS pair < ll , string >

typedef unsigned long long ull;
typedef long double ldbl;
typedef long long ll;
typedef double dbl;

const ldbl E = 2.71828182845904523536;
const ldbl pi = acos(-1);
const dbl eps = 1e-9;
const ll mod = 1e9+7;
const ll inf = 1LL<<30;

/** /////////// ** Code ** /////////// **/

const int N=200100;
vector<ii>adj[N];
map<int,ll>vis;
bool ok=1;

void DFS(int u,ll k){
    if(vis[u]){
        if(vis[u] != k) ok=0;
        return ;
    }

    vis[u]=k;
    for(auto v:adj[u])
        DFS(v.F,k+(ll)v.S);

}

int main()
{
    int n,m; scanf("%d%d",&n,&m);
    while(m--){
        int u,v,d; scanf("%d%d%d",&u,&v,&d);
        adj[u].push_back({v,d});
        adj[v].push_back({u,-1*d});
    }

    for(int i=1 ; i<=n ; i++){
        if(!vis[i]){
            DFS(i,0);
        }
    }

    if(ok) printf("Yes");
    else printf("No");

    return 0;
}