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 lint;
typedef pair<int, int> pi;
const int MAXN = 100005;

int n, m;
vector<pi> gph[MAXN];
bool vis[MAXN], bad;
int col[MAXN];

void dfs(int x, int c){
	if(vis[x]){
		if(col[x] != c) bad = 1;
		return;
	}
	vis[x] = 1;
	col[x] = c;
	for(auto &i : gph[x]) dfs(i.second, c + i.first);
}
int main(){
	scanf("%d %d",&n,&m);
	for(int i=0; i<m; i++){
		int s, e, x; scanf("%d %d %d",&s,&e,&x);
		gph[s].emplace_back(x, e);
		gph[e].emplace_back(-x, s);
	}
	for(int i=1; i<=n; i++){
		if(!vis[i]){
			dfs(i, 0);
		}
	}
	puts(!bad ? "Yes" : "No");
}