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;
using ll = long long;
ll n, m;
vector<pair<ll, ll>> v[100000];
bool used[100000];
ll koko[100000];
bool dfs(ll now){
	bool ret = 1;
	for(pair<ll, ll> next : v[now]){
		if(used[next.first]){
			if(koko[next.first] != koko[now] + next.second) return false;
		}else{
			used[next.first] = 1;
			koko[next.first] = koko[now] + next.second;
			ret &= dfs(next.first);
		}
	}
	return ret;
}
int main(){
	cin >> n >> m;
	for(ll i = 0; i < m; i++){
		ll l, r, d; cin >> l >> r >> d;
		v[--l].push_back(pair<ll, ll>(--r, d));
		v[r].push_back(pair<ll, ll>(l, -d));
	}
	bool ans = 1;
	for(ll i = 0; i < n; i++){
		if(!used[i]){
			used[i] = 1;
		}
		ans &= dfs(i);
	}
	cout << (ans ? "Yes\n" : "No\n");
	return 0;
}