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 n, p[100010], d[100010];
int getf(int x){
	if(p[x] == x)return x;
	getf(p[x]);
	int parent = p[x];
	p[x] = p[parent];
	d[x] = d[x] + d[parent];
	return p[x];
}

int main(){
	int n, m;
	scanf("%d%d" , &n , &m);
	for(int i = 1 ; i <= n ; i++)
		p[i] = i, d[i] = 0;

	bool invalid = false;

	for(int i = 1 ; i <= m ; i++){
		int l, r, dd;
		scanf("%d%d%d" , &l , &r , &dd);

		if(l == r){
			if(dd == 0)continue;
			else{
				invalid = true;
				continue;
			}
		}

		int pl = getf(l), pr = getf(r);

		if(pl == pr){
			if(d[r] - d[l] == dd){
				continue;
			}
			else{
				invalid = true;
			}
		}
		else{
			p[pr] = pl;
			d[pr] = dd - d[r] + d[l];
		}
	}

	if(invalid){
		puts("No");
		return 0;
	}

	puts("Yes");

	return 0;
}