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<cstdio>
using namespace std;

const int N = 1e5;
int fa[N+2];
long long fv[N+2];
int n,q;

int find_fa(int u)
{
	int i = u; long long t = 0ll;
	while(u!=fa[u]) {t += fv[u]; u = fa[u];}
	long long k = fv[i];
	fv[i] = t;
	while(i!=u)
	{
		t -= k;
		int j = fa[i];
		fa[i] = u;
		i = j;
		k = fv[i];
		fv[i] = t;
	}
	return u;
}

int main()
{
	scanf("%d%d",&n,&q); bool ans = true;
	for(int i=1; i<=n; i++) {fa[i] = i; fv[i] = 0ll;}
	for(int i=1; i<=q; i++)
	{
		int x,y; long long z; scanf("%d%d%lld",&x,&y,&z);
		int p = find_fa(x),q = find_fa(y);
		if(p!=q)
		{
			fa[q] = p; fv[q] = fv[x]+z-fv[y];
		}
		else
		{
			if(z!=fv[y]-fv[x]) {ans = false; break;}
		}
	}
	if(ans==true) printf("Yes\n");
	else printf("No\n");
	return 0;
}