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;
const int N = 1e5 + 5;
struct edge {
	int nxt, to, w;
} e[N << 2];
int n, m, cnt = 1;
int vis[N], h[N], val[N];
bool dfs(int u) {
	vis[u] = 1;
	for(int i = h[u]; i; i = e[i].nxt) {
		if(vis[e[i].to]) {
			if(val[e[i].to] != val[u] + e[i].w) {
				return false;
			}
		} else {
			val[e[i].to] = val[u] + e[i].w;
			if(!dfs(e[i].to)) {
				return false;
			} 
		}
	}
	return true;
}
void link(int u, int v, int w) {
	e[++cnt].nxt = h[u];
	h[u] = cnt;
	e[cnt].to = v;
	e[cnt].w = w; 
}
int main() {
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; ++i) {
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		link(u, v, w);
		link(v, u, -w);
	}
	for(int i = 1; i <= n; ++i) {
		if(!vis[i]) {
			if(!dfs(i)) {
				puts("No");
				return 0;
			}
 		}
	}
	puts("Yes");
	return 0;
}