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 cs const
#define pb push_back
using namespace std;
typedef long long ll;
cs int N = 1e5 + 50;
int n, q, fa[N], sz[N]; ll m;
int u[N], v[N], c;
int fnd(int x){ return x == fa[x] ? x : fa[x] = fnd(fa[x]); }
void mrg(int x, int y){
    x = fnd(x), y = fnd(y);
    if(x == y) return;
    fa[x] = y, sz[y] += sz[x];
}
int main(){
    #ifdef FSYolanda
    freopen("1.in","r",stdin);
    #endif
	scanf("%d%lld%d",&n,&m,&q);
    for(int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1;
    for(int i = 1, x, y, z; i <= q; i++){
        scanf("%d%d%d", &x, &y, &z); ++x, ++y;
        if(z == 0) mrg(x, y);
        else u[++c] = x, v[c] = y;
    } if(m == n - 1){
        if(c) return puts("No"), 0;
        else return puts("Yes"), 0;
    } 
    for(int i = 1; i <= c; i++){
        if(fnd(u[i]) == fnd(v[i])) 
        return puts("No"), 0;
    } ll lm = 0; int ct = 0;
    for(int i = 1; i <= n; i++)
    if(i == fnd(i)) lm += sz[i] - 1, ++ct;
    lm += 1ll * ct * (ct - 1) >> 1;
    if(m <= lm) puts("Yes");
    else puts("No"); return 0;
}