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<stdio.h>
int f[100010],v[100010];
int fa(int a)
{
    if(f[a]==a)
        return a;
    int t=f[a];
    f[a]=fa(f[a]);
    v[a]+=v[t];
    return f[a];
}
bool connect(int x,int y,int d)
{
    int fx=fa(x);
    int fy=fa(y);
    if(fx!=fy)
    {
        f[fy]=fx;
        v[fy]=v[x]-v[y]+d;
    }
    else if(v[y]-v[x]!=d)
        return 1;
    return 0;
}
int main()
{
    int n,m,h=0,l,r,d;
    scanf("%d%d",&n,&m);
    for(int j=1;j<=n;j++)
        f[j]=j;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&l,&r,&d);
        h+=connect(l,r,d);//true失败
    }
    if(!h)
        printf("Yes\n");
    else
        printf("No\n");
    return 0;
}