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;
typedef long long ll;

struct Edge{
	int to;
	ll D;
};

vector<Edge> G[100000];
ll X[100000];
int vis[100000];
bool ok = 1;

void dfs(int v){
	vis[v] = 1;
	for(auto e : G[v]){
		if(vis[e.to]){
			if(X[v] + e.D != X[e.to]) ok = 0;
		}
		else{
			X[e.to] = X[v] + e.D;
			dfs(e.to);
		}
	}
}


int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	int N, M;
	cin >> N >> M;
	for(int i = 0; i < M; i++){
		int L, R, D;
		cin >> L >> R >> D;
		L--, R--;
		G[L].push_back({ R, D });
		G[R].push_back({ L, -D });
	}

	for(int i = 0; i < N; i++){
		if(vis[i]) continue;
		X[i] = 0;
		dfs(i);
	}

	if(ok){
		cout << "Yes" << endl;
	}
	else{
		cout << "No" << endl;
	}
}