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>
using namespace std;
int father[100500];
int dis[100500];
int find_father(int x) {
	if(father[x]==x)return x;
	int a1=father[x];
	int a2=find_father(father[x]);
	father[x]=a2;
	dis[x]=dis[x]+dis[a1];
	return a2;
}
int find_dis(int x) {
	if(father[x]==x)return 0;
	return find_dis(father[x])+dis[x];
}
int main(void) {
	int num,cas,L,R,D;
	cin>>num>>cas;
	for(int i=0; i<=num; i++) {
		father[i]=i;
		dis[i]=0;
	}
	while(cas--) {
		cin>>L>>R>>D;
		int fl=find_father(L);
		int fr=find_father(R);
		if(fl==fr) {
			if(dis[L]+D!=dis[R]) {
				cout<<"No"<<endl;
				return 0;
			}
		} else {
			father[fr]=fl;
			dis[fr]=find_dis(L)-find_dis(R)+D;
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}