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 x first
#define y second
#define pb push_back
#define Set(i, v) memset(i, v, sizeof i)
#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

typedef long long LL;

using namespace std;

const int N = 1e5 + 10;

int n, m;
int X[N];
bool vis[N];
vector<int> G[N], w[N];

void DFS(int o){
	vis[o] = true;
	For(i, 0, (int)G[o].size() - 1){
		int v = G[o][i], d = w[o][i];
		if(vis[v] && X[v] != X[o] + d){
			puts("No");
			exit(0);
		}
		else if(!vis[v]){
			X[v] = X[o] + d;
			DFS(v);
		}
	}
}

int main(){

	scanf("%d%d", &n, &m);
	For(i, 1, m){
		int u, v, d;
		scanf("%d%d%d", &u, &v, &d);
		G[u].pb(v), G[v].pb(u);
		w[u].pb(d), w[v].pb(-d);
	}

	For(i, 1, n) if(!vis[i]) DFS(i);
	puts("Yes");

	return 0;
}