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 x first
#define y second

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n, m, q, comp;
int par[100100];

void init() {
    for (int i=0;i<n;i++) par[i] = i;
}
int fin(int a) {
    return par[a] = (par[a]==a?a:fin(par[a]));
}
void uni(int a, int b){
    a = fin(a); b = fin(b);
    if (a==b) return;
    comp--;
    par[fin(a)] = fin(b);
}
void pung() {
    puts("No");
    exit(0);
}
void print(bool t) {
    puts(t?"Yes":"No");
    exit(0);
}

int main() {
    scanf("%d%d%d",&n,&m,&q);
    init(); comp = n;
    vector<pii> vec;
    for (int i=0;i<q;i++) {
        int a, b, c;
        scanf("%d%d%d",&a,&b,&c);
        if (c==0) uni(a,b);
        else {
            vec.push_back({a,b});
        }
    }
    for (pii &v : vec) {
        if (fin(v.first)==fin(v.second)) pung();
    }
    if (comp==1) {
        print(m==n-1);
    }
    if (comp==2) {
        print(vec.empty());
    }
    if (vec.empty()) {
        print(m<=n-comp+1LL*comp*(comp-1)/2);
    }
    print(n<=m&&m<=n-comp+1LL*comp*(comp-1)/2);

	return 0;
}