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;
typedef long long int ll;
#define N 100005
ll n, m, x, y, z, f[N], d[N], r[N], s;
bool ish(ll q)
{
	return q >= 0;
}
void go(ll w)
{
	ll q = w, e = f[q];
	while (e != f[e])
	{
		d[q] += d[e];
		f[q] = f[e];
		e = f[e];
	}
}
void merge()
{
	ll q = f[x], w = f[y], t = d[x] - d[y] + z;
	if (r[q] < r[w])
		swap(q, w), t = -t;
	f[w] = q;
	d[w] = t;
	r[q] += r[w];
}
int main()
{
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= n; i++)
		f[i] = i, r[i] = 1;
	for (int i = 0; i < m; i++)
	{
		scanf("%lld%lld%lld", &x, &y, &z);
		go(x); go(y);
		/*for (int i = 1; i <= n; i++)
			cout << i << " " << d[i] << endl;*/
		if (f[x] == f[y])
		{
			if(d[y]-d[x]!=z)
			{
				cout << "No\n";
				return 0;
			}
		}
		else
			merge();
	}

	cout << "Yes\n";
	return 0;
}