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, m, d[100005], p[100005];
vector<int> c[100005];

int Find (int X) {
	if(p[X] == X) return X;
	return p[X] = Find(p[X]);
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) {
		p[i] = i;
		c[i].push_back(i);
	}
	while(m--) {
		int L, R, C;
		scanf("%d%d%d",&L,&R,&C);
		int LL = Find(L), RR = Find(R);
		if(LL == RR) {
			if(d[R] - d[L] != C) {
				puts("No");
				return 0;
			}
			continue;
		}
		if(c[LL].size() < c[RR].size()) {
			swap(LL, RR);
			swap(L, R);
			C *= -1;
		}
		int D = d[L] + C - d[R];
		for(auto &T : c[RR]) {
			c[LL].push_back(T);
			d[T] += D;
		}
		p[RR] = LL;
		c[RR].clear();
	}
	puts("Yes");
}