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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define EPS (1e-10)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

vector<P>E[100000];
bool used[100000];
int a[100000];

void dfs(int v){
	used[v]=true;
	for(auto p:E[v]){
		if(used[p.second]){
			if(a[v]+p.first!=a[p.second]){
				puts("No");exit(0);
			}
		}
		else{
			a[p.second]=a[v]+p.first;
			dfs(p.second);
		}
	}
}
signed main(){
	int n,m;scanf("%d%d",&n,&m);
	rep(i,m){
		int l,r,d;scanf("%d%d%d",&l,&r,&d);l--;r--;
		E[l].push_back(P(d,r));
		E[r].push_back(P(-d,l));
	}
	rep(i,n){
		if(!used[i]){
			dfs(i);
		}
	}
	puts("Yes");
}