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>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX=1e6;
int father[MAX],v[MAX],flag,t;
int find(int x)
{
    if(father[x]==x)return x;
    t=find(father[x]);
    v[x]+=v[father[x]];
    father[x]=t;
    return father[x];
}
void work(int x,int y,int w)
{
    int p=find(x),q=find(y);
    if(p!=q)
    {
        father[p]=q;
        v[p]=v[y]-v[x]-w;
    }
    else if(v[y]-v[x]!=w)flag=1;
}
int main()
{
    int w,n,m;
    memset(v,0,sizeof(v));
    flag=0;
    scanf("%d%d",&n,&m);
    for(int i=0; i<=n; i++)father[i]=i;
    for(int i=1; i<=m; i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        work(x,y,z);
    }
    if(flag)printf("No\n");
    else printf("Yes\n");
    return 0;
}