#include<bits/stdc++.h>
#define pb push_back
using namespace std;
int n,m,x,y,z;
int ans;
int du[100001],bz[100001];
struct yu
{
    int x,y;
    bool operator<(const yu &a)const
    {
        return a.x<x||(a.x==x&&a.y<y);
    }
};
vector<yu>e[100001];
map<yu,int>A;
yu L=(yu){0,0};
int main()
{
    scanf("%d %d",&n,&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d %d %d",&x,&y,&z);
        if (x>y)swap(x,y);z++;
        if (A.count((yu){x,y}))A[(yu){x,y}]+=z;else A[(yu){x,y}]=z;
    }
    ans=n;
    for (auto [i,j]:A)
    {
        if (j==3)ans++;
        e[i.x].pb((yu){i.y,j}),du[i.y]++;
        e[i.y].pb((yu){i.x,j}),du[i.x]++;
    }
    for (auto [i,j]:A)if (j==3)
    {
        int x=i.x,y=i.y;
        if (du[x]!=2||du[y]!=2)continue;
        yu o=L,p=L;
        for (yu I:e[x])if (I.x!=y)o=(yu){I.x,I.y};
        for (yu I:e[y])if (I.x!=x)p=(yu){I.x,I.y};
        if (o.x==p.x&&o.y!=p.y)ans++;
    }
    int ank=0;
    for (auto [i,j]:A)if (j==3)
    {
        int x=i.x,y=i.y;
        if (du[x]!=2||du[y]!=2)continue;
        yu o=L,p=L;
        for (yu I:e[x])if (I.x!=y)o=(yu){I.x,I.y};
        for (yu I:e[y])if (I.x!=x)p=(yu){I.x,I.y};
        if (o.x!=p.x&&o.y!=p.y)
        {
            if (du[o.x]!=2||du[p.x]!=2)continue;
            yu u=L,v=L;
            for (yu I:e[o.x])if (I.x!=x)u=(yu){I.x,I.y};
            for (yu I:e[p.x])if (I.x!=y)v=(yu){I.x,I.y};
            if (u.x==p.x&&v.x==o.x&&u.y==3&&v.y==3)ank++;
        }
    }
    int ano=0;
    for (int i=1;i<=n;i++)if (du[i]==3)
    {
        int l=0,r=0;
        if (e[i][0].y+e[i][1].y+e[i][2].y==4)l++;else if (e[i][0].y+e[i][1].y+e[i][2].y==5)r++;
        vector<yu>b;
        bz[i]=1;
        for (yu j:e[i])b.pb(j),bz[j.x]=1;
        int BZ=1;
        if (du[b[0].x]!=3||du[b[1].x]!=3||du[b[2].x]!=3)continue;
        for (yu k:b)
        {
            for (yu w:e[k.x])
            {
                if (bz[w.x]==0)
                {
                    BZ=0;
                    break;
                }
            }
            int ii=k.x;
            if (e[ii][0].y+e[ii][1].y+e[ii][2].y==4)l++;else if (e[ii][0].y+e[ii][1].y+e[ii][2].y==5)r++;
            if (BZ==0)break;
        }
        if (l==2&&r==2&BZ)ano++;

        bz[i]=0;
        for (yu j:e[i])bz[j.x]=0;
    }
    //printf("%d\n",ano);
    printf("%d\n",ans+ank/2+ano/4);
}