#include<iostream>
#include<stdio.h>
using namespace std;
const int N = 50010;
int p[N], d[N];  //d[i]是i到根节点的距离，距离为0就是根节点本身，为1是吃根的，2是吃1且被根吃的
int n, m, res;
int find(int x)
{
    if(p[x] != x)
    {
        int t = p[x];
        p[x] = find(p[x]);
        d[x] += d[t];
    }
    return p[x];
}
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++) 
    {
        p[i] = i; d[i] = 0;
    }
    while(m --)
    {
        int k, a, b;
        scanf("%d%d%d", &k, &a, &b);
        if(a > n || b > n)
        {
            res ++;
            continue;
        }
        int px = find(a);
        int py = find(b);
        if(k == 1)
        {
            if(px == py && (d[a] - d[b]) % 3) res ++;  //如果ab在同一集合而且不是同类关系的话
            else if(px != py)                           //等价于(d[a] - d[b]) % 3 != 0
            {
                p[px] = py;
                d[px] = d[b] - d[a]; 
            }
        }
        else 
        {                                                   
            if(px == py && (d[a] - d[b] - 1) % 3) res ++;  //等价于(d[a] - d[b] - 1) % 3 != 0
            else if(px != py)   
            {
                p[px] = py;
                d[px] = d[b] + 1 - d[a];
            }

        }
    }
    cout << res;
    return 0;
}