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 <iostream>
using namespace std;
typedef long long ll;
ll n,m,l,r,d,fa[200001],ra[200001],diff_weight[200001];
ll find(ll x){
    if(fa[x]==x)return x;
    ll xx=find(fa[x]);
    diff_weight[x]+=diff_weight[fa[x]];
    return fa[x]=xx;
}
ll weight(ll x){
    find(x);
    return diff_weight[x];
}
bool uni(ll x,ll y,ll w){
    w+=weight(x),w-=weight(y);
    x=find(x);
    y=find(y);
    if(x==y)return false;
    if(ra[x]<ra[y]){swap(x,y);w=-w;}
    fa[y]=x;
    if(ra[x]==ra[y])ra[x]++;
    diff_weight[y]=w;
    return true;
}
int main(void){
    cin>>n>>m;
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=0;i<m;i++){
        cin>>l>>r>>d;
        if(find(l)!=find(r))uni(l,r,d);
        else{
        if(weight(r)-weight(l)!=d){cout<<"No"<<endl;return 0;}
        }
    }
    cout<<"Yes"<<endl;
}