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 <iostream>
#include <cstdio>
#include <vector>

using namespace std;


typedef pair<int,int> pii;

const int N = 1e5 + 5;

vector<int> g[N];
bool used[N];
int comp[N];

void dfs(int v)
{
    used[v] = true;
    for (int u : g[v])
    {
        if (used[u]) continue;
        comp[u] = comp[v];
        dfs(u);
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);


    int n, q;
    long long m;
    scanf("%d%lld%d", &n, &m, &q);

    vector<pii> edges;
    while(q--)
    {
        static int u, v, type;
        scanf("%d%d%d", &u, &v, &type);
        if (type)
            edges.push_back({u, v});
        else
        {
            g[u].push_back(v);
            g[v].push_back(u);
        }
    }

    int cntComp = 0;
    for (int v = 0; v < n; ++v)
    {
        if (!used[v])
        {
            comp[v] = cntComp++;
            dfs(v);
        }
    }

    for (pii e : edges)
    {
        if (comp[e.first] == comp[e.second])
        {
            printf("No");
            return 0;
        }
    }

    m -= n - cntComp;

    if (!edges.empty() && cntComp == 2)
    {
        printf("No");
        return 0;
    }
    if (cntComp - (int)edges.empty() <= m && m <= cntComp * 1LL * (cntComp - 1) / 2)
        printf("Yes");
    else
        printf("No");
}