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:
//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>
using namespace std;
#define pb push_back//also for STRING

#define MAX_N 100000

struct edge{
	int to,cost;
};

int d[MAX_N];
vector<edge> g[MAX_N];
bool f[MAX_N];

bool dfs(int v){
	f[v]=true;
	for(int j=0;j<g[v].size();++j){
		int w=g[v][j].to;
		if(f[w]){
			if(d[w]!=d[v]+g[v][j].cost) return false;
			continue;
		}
		d[w]=d[v]+g[v][j].cost;
		if(!dfs(w)) return false;
	}
	return true;
}

int main(){
	int n,m;
	scanf("%d %d",&n,&m);
	while(m-- > 0){
		int l,r,d;
		scanf("%d %d %d",&l,&r,&d);
		g[--l].pb((edge){--r,d});
		g[r].pb((edge){l,-d});
	}
	fill(f,f+n,false);
	for(int v=0;v<n;++v){
		if(!dfs(v)){
			printf("No\n");
			return 0;
		}
	}
	printf("Yes\n");
	return 0;
}