Problem: Snuke's mother gave Snuke an undirected graph consisting of N vertices numbered 0 to N-1 and M edges. This graph was connected and contained no parallel edges or self-loops.

One day, Snuke broke this graph. Fortunately, he remembered Q clues about the graph. The i-th clue (0 \leq i \leq Q-1) is represented as integers A_i,B_i,C_i and means the following:

* If C_i=0: there was exactly one simple path (a path that never visits the same vertex twice) from Vertex A_i to B_i.
* If C_i=1: there were two or more simple paths from Vertex A_i to B_i.



Snuke is not sure if his memory is correct, and worried whether there is a graph that matches these Q clues. Determine if there exists a graph that matches Snuke's memory.

Constraints

* 2 \leq N \leq 10^5
* N-1 \leq M \leq N \times (N-1)/2
* 1 \leq Q \leq 10^5
* 0 \leq A_i,B_i \leq N-1
* A_i \neq B_i
* 0 \leq C_i \leq 1
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M Q
A_0 B_0 C_0
A_1 B_1 C_1
\vdots
A_{Q-1} B_{Q-1} C_{Q-1}


Output

If there exists a graph that matches Snuke's memory, print `Yes`; otherwise, print `No`.

Examples

Input

5 5 3
0 1 0
1 2 1
2 3 0


Output

Yes


Input

4 4 3
0 1 0
1 2 1
2 3 0


Output

No


Input

10 9 9
7 6 0
4 5 1
9 7 0
2 9 0
2 3 0
4 1 0
8 0 0
9 1 0
3 0 0


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define dbug(x) cout<<#x<<"="<<x<<endl
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=(2e5)+10;
int n,q,vis[maxn];
ll m,s,cnt;
struct node {
	int x,y,z;
} d[maxn];
struct Seg {
	int fa[maxn],sz[maxn];
	void init() {
		for (int i=1;i<=n;i++) fa[i]=i,sz[i]=1;
	}
	int find(int x) {
		if (fa[x]==x) return x;
		return fa[x]=find(fa[x]);
	}
	void connect(int x,int y) {
		x=find(x); y=find(y);
		if (x==y) return;
		fa[x]=y; sz[y]+=sz[x];
	}
} A,B;
int main() {
	//freopen("1.txt","r",stdin);
	read(n); read(m); read(q);
	int x,y,z,flag=0; A.init();
	for (int i=1;i<=q;i++) {
		read(x); read(y); read(z);
		x++; y++;
		d[i].x=x; d[i].y=y; d[i].z=z;
		if (!z) A.connect(x,y);
	}
	for (int i=1;i<=q;i++) {
		x=d[i].x; y=d[i].y; z=d[i].z;
		if (!z) continue;
		x=A.find(x); y=A.find(y);
		if (x==y) { printf("No\n"); return 0; }
		flag=1;
	}
	for (int i=1;i<=n;i++)
		if (A.find(i)==i) cnt++;
	if (m==n-1) {
		if (flag) printf("No\n");
		else printf("Yes\n");
		return 0;
	}
	if (m-(n-cnt)<=cnt*(cnt-1)/2) printf("Yes\n");
	else printf("No\n");
	return 0;
}