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;

#define N 600000
#define INF 0x3f3f3f3f

int n,m,x,y,z,tot,f[N],head[N];
struct edge{int v,l,nxt;}e[N];

void add(int x,int y,int z){e[++tot].v=y; e[tot].l=z; e[tot].nxt=head[x]; head[x]=tot;}

void dfs(int u){
	for (int i=head[u],v;i;i=e[i].nxt)
		if (f[v=e[i].v]==INF){
			f[v]=f[u]+e[i].l;
			dfs(v);
		}
		else
			if (f[v]!=f[u]+e[i].l){puts("No"); exit(0);}
}

int main(){
	scanf("%d%d",&n,&m);
	memset(f,0x3f,sizeof f);
	while (m--){
		scanf("%d%d%d",&x,&y,&z);
		add(x,y,z); add(y,x,-z);
	}
	for (int i=1;i<=n;++i)
		if (f[i]==INF){f[i]=0; dfs(i);}
	puts("Yes");
	
	return 0;
}