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 BUF (200005)
#define ll long long 

int n,m;
vector<pair<int,int>> e[BUF];

int used[BUF];
int pos[BUF];

int dfs(int nd){
	
	vector<int> next;
	for(auto x:e[nd]){
		if(used[x.first]==0){
			used[x.first]=1;
			pos[x.first]=pos[nd]+x.second;
			next.push_back(x.first);
		}else{
			if(pos[x.first]!=pos[nd]+x.second) return -1;
		}
	}
	int ret=0;
	for(auto x:next){
		if(dfs(x)<0)ret = -1;
	}
	return ret;

}

int main(void){
	cin >> n >> m;

	for(int i=1;i<=m;i++){
		pair<int,int> p;
		int l,r,d;
		cin >> l >> r >> d;
		p.first=r;
		p.second=d;
		e[l].push_back(p);
		p.first=l;
		p.second=-d;
		e[r].push_back(p);
	}

	for(int i=1;i<=n;i++){
		if(used[i]==0){
			if(dfs(i)<0){
				cout << "No\n";
				return 0;
			}
		}
	}

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