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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
typedef long long ll;
typedef long double lb;
#define int long long
#define double long double
#define endn "\n"
typedef pair<int,int> P;
template<class T> inline bool chmin(T& a, T b){if(a>b){a = b;return 1;}return 0;}
template<class T> inline bool chmax(T& a, T b){if(a<b){a = b;return 1;}return 0;}
const int MOD = 1000000007;
//const int MOD = 998244353;
template<class T> inline int add(T& a, T b, T M = MOD){a=(a+M)%M;b=(b+M)%M;a=(a+b)%M;return a;};
template<class T> inline int mul(T& a, T b, T M = MOD){if(a>=M)a%=M;if(b>=M)b%=M;a*=b;if(a>=M)a%=M;return a;};
const ll INF = 1e16;
const double EPS = 1e-10;
const double PI = 3.141592653589793;
const string abc="abcdefghijklmnopqrstuvwxyz";
const string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const pair<int,int> fd[] = {make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)}; 

////////////////////////////////////////////////////////////////////
///////////////////////___modpow___////////////////////
ll modpow(ll a,  ll n, const ll &MOD = MOD){
  ll ret = n == 0 ? 1 : modpow(a, n/2, MOD);
  (ret *= ret);
  if(n%2)((ret *= a));
  return ret;
}
///////////////////////___modinv___////////////////////
ll modinv(ll d, const ll &MOD = MOD){
  return modpow(d, MOD-2, MOD);
}

vector<ll> fac, finv, inv;
void COMinit(const int MAX = 5000000, const ll &MOD = MOD){
  fac.resize(MAX);finv.resize(MAX);inv.resize(MAX);
  fac[0] = fac[1] = finv[0] = finv[1] = inv[1] = 1;
  repd(i,2,MAX){
    fac[i] = fac[i-1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
    finv[i] = finv[i-1] * inv[i] % MOD;
  }
}
ll COM(int n, int k){
  if(n < k)return 0;
  if(n < 0 || k < 0)return 0;
  return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD;
}


////////////////////////////////////////////////////////////////////

int n,m;vector<P> g[110000];
int dist[110000];bool c;
int dfs(int x){
  rep(i,g[x].size()){
    int to = g[x][i].first, cost = g[x][i].second;
    if(dist[to] == INF){
      dist[to] = dist[x]+cost;
      dfs(to);
    }else if(dist[to] != dist[x]+cost){
      c = false;
    }
  }
}
signed main(){
  cin>>n>>m;
  rep(i,m){
    int l,r,d;cin>>l>>r>>d;l--;r--;
    g[l].push_back(P(r,d));
    g[r].push_back(P(l,-d));
  }
  c = true;
  rep(i,110000)dist[i] = INF;
  rep(i,n)if(dist[i] == INF)dfs(i);
  cout << (c ? "Yes" : "No") << endl;
}