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 MAXN 200100
#define pii pair<int, int>
#define pb push_back
typedef long long ll;

using namespace std;

int n, m, l[MAXN], r[MAXN], d[MAXN], p[MAXN];

vector<pii> g[MAXN];
bool vis[MAXN];

void dfs(int x, int pos){
	if(vis[x]) return;
	vis[x] = true;
	p[x] = pos;
	for(auto i : g[x]){
		dfs(i.first, i.second + pos);
	}
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    cin >> n >> m;
    for(int i = 0; i < m; i++){
    	cin >> l[i] >> r[i] >> d[i];
    	l[i]--;
    	r[i]--;
    	g[l[i]].pb({r[i], d[i]});
    	g[r[i]].pb({l[i], -d[i]});
    }
    for(int i = 0; i < n; i++)
    	dfs(i, 0);
    for(int i = 0; i < m; i++){
    	if(p[r[i]] - p[l[i]] != d[i]){
    		cout << "No";
    		return 0;
    	}
    }
    cout << "Yes";
}