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>
#define MAX_N 100005
#define pb push_back
using namespace std;
struct node{
	int y,cost;
	bool operator < (node d) const{
		return cost<d.cost;
	}
}drg[MAX_N];
int n,m,dst[MAX_N];
bool ans;
vector<node> net[MAX_N];
void dfs(int x){
	int sz=net[x].size();
	node y;
	while(sz--){
		y=net[x][sz];
		if(dst[y.y]){
			if(dst[x]+y.cost!=dst[y.y])	ans=true;
			continue;
		}
		dst[y.y]=dst[x]+y.cost;
		dfs(y.y);
	}
}
int main(){
	int i,x;
	node p;
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)	drg[i].y=i;
	for(i=0;i<m;i++){
		scanf("%d %d %d",&x,&p.y,&p.cost);
		net[x].pb(p);
		drg[p.y].cost++;
	}
	sort(drg+1,drg+n+1);
	for(i=1;i<=n;i++){
		if(dst[drg[i].y])	continue;
		dst[drg[i].y]=1;
		dfs(drg[i].y);
	}
	printf("%s\n",(ans)?"No":"Yes");
	return 0;
}