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 MAXN = 100010;

typedef long long ll;
const int mod = 1e9 + 7;

#define X first
#define Y second

vector<pair<int, int> > E[MAXN];
bool flag = true;
bool vis[MAXN];
int d[MAXN];

void dfs(int x)
{
	for (auto e: E[x])
	{
		int y = e.X, z = e.Y;
		if (!vis[y])
		{
			vis[y] = true;
			d[y] = d[x] + z;
			dfs(y);
		}
		else if (d[y] != d[x] + z) flag = false;
	} 
}

int main()
{
	//freopen("input", "r", stdin);

	int n, m, x, y, z;
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= m; ++i)
	{
		scanf("%d %d %d", &x, &y, &z);
		E[x].push_back(make_pair(y, z));
		E[y].push_back(make_pair(x, -z));
	}
	
	for (int i = 1; i <= n; ++i)
		if (!vis[i]) vis[i] = true, d[i] = 0, dfs(i);
	
	puts(flag? "Yes":"No");

    return 0;
}