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>
typedef long long ll;

using namespace std;

const int N=1e5+5;
const int INF=0x3f3f3f;

int par[N];
int v[N];

int Find(int x){
    int temp=par[x];
    if(par[x]!=x)   par[x]=Find(par[x]),v[x]+=v[temp];
    return par[x];
}

int main(void){
    int n,m;
    cin >>n >>m;
    for(int i=1;i<=n;i++)   par[i]=i;
    bool flag=true;
    for(int i=1;i<=m;i++){
        int l,r,d;
        scanf("%d%d%d",&l,&r,&d);
        int rootl=Find(l);
        int rootr=Find(r);
        if(rootr==rootl){
            if(v[r]-v[l]!=-d)   flag=false;
        }
        else{
            par[rootl]=rootr;
            v[rootl]=d+v[r]-v[l];
        }
    }
    if(flag)    cout <<"Yes"<<endl;
    else    cout<<"No"<<endl;
}