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>
const long long maxn=1e5+100;
using namespace std;
long long n,m,k,t,f[maxn],x,y,d[maxn];bool le;
int getf(int u)
{
    int k;
    if(f[u]==u)
        return u;
    k=f[u];
    f[u]=getf(f[u]);
    d[u]=d[u]+d[k];
    return f[u];
}
void jia(int x,int y,int k)
{
    int i,j;
    i=getf(x);
    j=getf(y);
    f[j]=i;
    d[j]=d[x]-d[y]-k;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for(int i=1;i<=n;++i)
        f[i]=i;
    for(int i=1;i<=m;++i)
    {
        scanf("%lld%lld%lld",&x,&y,&k);
        if(getf(x)!=getf(y))
            jia(y,x,k);
        else
            if(d[y]-d[x]!=k)
        {
            le=1;
        }
    }
    if(le)
        printf("No\n");
    else
        printf("Yes\n");
    return 0;
}