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<iostream>
using namespace std;
const int maxn = 100005;
int f[maxn], sum[maxn];
int find(int a)
{
	if (f[a] != a)
	{
		int pre = f[a];
		f[a] = find(f[a]);
		sum[a] = sum[a] + sum[pre];
	}
	return f[a];
}
void Union(int u, int v, int rootu, int rootv, int val)
{
	f[rootu] = rootv;
	sum[rootu] = -sum[u] + val + sum[v];
}

int main()
{
	int N, M;
	scanf("%d %d", &N, &M);
	for (int i = 1;i <= N;i++)f[i] = i,sum[i]=0;
	int u, v, rootu, rootv, dist;
	bool flag = true;
	for (int i = 1;i <= M;i++)
	{
		scanf("%d %d %d", &u, &v, &dist);
		rootu = find(u);
		rootv = find(v);
		if (rootu != rootv)
		{
			Union(u, v, rootu, rootv, dist);
		}
		else
		{
			if (sum[u] - sum[v] != dist)
				flag = false;
		}
	}
	if (flag)puts("Yes");
	else puts("No");
	return 0;
}