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>
#include<algorithm>
#include<vector>
using namespace std;
long long D[101000];
int ck[101000], n, m;
vector<int>E[101000], L[101000];
void DFS(int a, long long d) {
	int i;
	ck[a] = 1;
	D[a] = d;
	for (i = 0; i < E[a].size(); i++) {
		if (!ck[E[a][i]])DFS(E[a][i], D[a] + L[a][i]);
	}
}
int main() {
	int i, a, b, c;
	scanf("%d%d", &n, &m);
	for (i = 1; i <= m; i++) {
		scanf("%d%d%d", &a, &b, &c);
		E[a].push_back(b);
		L[a].push_back(c);
		E[b].push_back(a);
		L[b].push_back(-c);
	}
	for (i = 1; i <= n; i++) {
		if (!ck[i]) {
			DFS(i, 0);
		}
	}
	for (i = 1; i <= n; i++)for (int j = 0; j < E[i].size(); j++) {
		if (D[E[i][j]] != D[i] + L[i][j]) {
			printf("No\n");
			return 0;
		}
	}
	printf("Yes\n");
}