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<cstdio>
#include<algorithm>
#include<cstring>
#define rep(i, s, t) for(i = s; i <= t; ++i)
#define dep(i, s, t) for(i = s; i >= t; --i)

using namespace std;

const int N = 1e5 + 10;
int n, m; bool ans = 1;
int pos[N];
struct edge{int ed, v, ne;} e[N<<1]; int he[N], in[N], en;

void ins(int a, int b, int c) {
	e[++en].ed = b; e[en].v = c; e[en].ne = he[a]; he[a] = en; ++in[b];
}

void dfs(int u) {
	for(int i = he[u], v; i; i = e[i].ne) {
		v = e[i].ed;
		if(!pos[v]) {
			pos[v] = pos[u] + e[i].v;
			dfs(v);
		} else 
			if(pos[u] + e[i].v != pos[v]) ans = 0;
	}
}

int main() {
	int i, x, y, z;
	scanf("%d%d", &n, &m);
	rep(i, 1, m) {scanf("%d%d%d", &x, &y, &z); ins(x, y, z);}
	rep(i, 1, n) if(!in[i]) {pos[i] = 1; dfs(i);}
	rep(i, 1, n) if(!pos[i]) ans = 0;
	if(ans) puts("Yes"); else puts("No");
	return 0;
}