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 <vector>

using namespace std;

vector<pair<int, int> > g[100001];
int v[100001];
int d[100001];

int f(int x, int y) {
	if (v[x]) return d[x] != y;
	v[x] = 1;
	d[x] = y;
	for (auto t : g[x]) if (f(t.first, y + t.second)) return 1;
	return 0;
}

int main() {
	int i, j, k, n, m;
	scanf("%d%d", &n, &m);
	while (m--) {
		scanf("%d%d%d", &i, &j, &k);
		g[i].emplace_back(j, k);
		g[j].emplace_back(i, -k);
	}
	for (i = 1; i <= n; i++) if (!v[i] && f(i, 0)) {
		puts("No");
		return 0;
	}
	puts("Yes");
	return 0;
}