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;

const int N = 1e5 + 10;

int p[N], rec[N];

int find(int x){
	if (p[x] == x) return x;
	int fa = find(p[x]);
	rec[x] += rec[p[x]];
	return p[x] = fa;
}

bool same(int x, int y){
	return find(x) == find(y);
}

void unite(int x, int y, int val){
	int fx = find(x), fy = find(y);
	p[fx] = fy;
	rec[fx] = rec[y] - rec[x] + val;
}

int main(){
	int n, m, x, y, d;
	scanf("%d %d",&n,&m);
	for (int i = 1; i <= n; i++){
		p[i] = i;
		rec[i] = 0;
	}
	for (int i = 1; i <= m; i++){
		scanf("%d %d %d",&x,&y,&d);
		if (same(x, y)){
			if (rec[x] - rec[y] != d){
				printf("No\n");
				return 0;
			}
		}
		else unite(x, y, d);
	}
	printf("Yes\n");
	return 0;
}