Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> pi;
typedef vector<int> vi;
typedef vector<pi> vpi;
#define f first
#define pb push_back
#define s second
#define mp make_pair

int d[100100];
vector<pi> aList[100100];
int N,M,a,b,c;
int INF = 1e9;

void dfs(int x){
  for (auto i : aList[x]){
    int nxt = d[x] + i.s;
    if (d[i.f] == -INF){
      d[i.f] = nxt;
      dfs(i.f);
    }else{
      if (nxt != d[i.f]){
        cout<<"No";
        exit(0);
      }
    }
  }
}

int main(){
  cin >> N >> M;
  for (int i=0;i<M;++i){
    cin>>a>>b>>c;
    aList[a].pb(mp(b,c));
    aList[b].pb(mp(a,-c));
  }
  for (int i=1;i<=N;++i)d[i] = -INF;
  for (int i=1;i<=N;++i){ 
    if (d[i] != -INF)continue;
    d[i] = d[i-1];
    dfs(i);
  }
  cout<<"Yes";
}