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;
#define fi first
#define se second
#define mp make_pair
typedef pair<int, int> ii;
const int N = 1e5 + 5;
const int M = 2e5 + 5;
int n, m;
int l[M], r[M], d[M], mark[N];
long long sum[N];
vector<ii> adj[N];

void dfs(int u) {
	mark[u] = 1;
	for (ii v : adj[u]) {
		if (!mark[v.fi]) {
			sum[v.fi] = sum[u] + v.se;
			dfs(v.fi);
		}
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i ++) {
		scanf("%d%d%d", &l[i], &r[i], &d[i]);
		adj[l[i]].push_back(mp(r[i], d[i]));
		adj[r[i]].push_back(mp(l[i], -d[i]));
	}
	for (int u = 1; u <= n; u ++) {
		if (mark[u] == 0) {
			dfs(u);
		}
	}
	for (int i = 1; i <= m; i ++) {
		if (sum[r[i]] - sum[l[i]] != d[i]) {
			printf("No\n");
			exit(0);
		}
	}
	printf("Yes\n");
}