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;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 100000;
map<int, int> mp[N + 1];
int pa[N + 1];

int main(){
	int n, m;
	scanf("%d%d", &n, &m);
	f(i, 1, n + 1)pa[i] = i;
	f(i, 1, n + 1)mp[i][i] = 0;
	while (m--){
		int l, r, d;
		scanf("%d%d%d", &l, &r, &d);
		if (pa[l] == pa[r]) { if (mp[pa[l]][r] - mp[pa[l]][l] != d) { printf("No\n"); return 0; } }
		else {
			int pl = pa[l], pr = pa[r];
			if (mp[pl].size() > mp[pr].size())swap(pl, pr), swap(l, r), d = -d;
			int z = mp[pr][r] - d - mp[pl][l];
			for (map<int, int>::iterator it = mp[pl].begin(); it != mp[pl].end(); mp[pl].erase(it++)){
				pa[it->first] = pr;
				mp[pr][it->first] = it->second + z;
			}
		}
	}
	printf("Yes\n");
}