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;
vector<pair<int, int> > adjlist[100005];
int inc[100005];
bitset<100005> vis;
int dist[100005];
bool flag=true;

void dfs(int v, int d) {
	if (dist[v]==-1) dist[v]=d;
	else if (dist[v]!=d) flag=false;
	//printf("%d %d\n", v, dist[v]);
	vis[v]=true;
	for (auto it:adjlist[v]) {
		if (!vis[it.first]) dfs(it.first, d+it.second);
		if (d+it.second!=dist[it.first]) flag=false;
	}
}

int main() {
	memset(dist,-1,sizeof(dist));
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i=0; i<m; i++) {
		int l,r,d;
		scanf("%d%d%d", &l, &r, &d);
		l--; r--;
		adjlist[l].push_back({r,d});
		inc[r]++;
	}
	for (int i=0; i<n; i++) {
		if (vis[i]||inc[i]) continue;
		dfs(i,0);
	}
	for (int i=0; i<n; i++) flag &= vis[i];
	printf("%s\n", flag?"Yes":"No");
}