Rt.

此 hack 可以卡掉一些能通过原数据的玄学做法（比如我的）

大致就是这样造。

#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define rep(i,x,y) for(register int i=x;i<=y;i++)
#define rep1(i,x,y) for(register int i=x;i>=y;--i)
#define int long long
#define fire signed
#define il inline
template<class T> il void print(T x) {
	if(x<0) printf("-"),x=-x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + '0');
}
template<class T> il void in(T &x) {
    x = 0; char ch = getchar();
    int f = 1;
    while (ch < '0' || ch > '9') {if(ch=='-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
    x *= f;
}
int T=1;
void solve() {
	int n=300000,m=300000;
	cout<<1<<' '<<n<<' '<<m<<endl;
	int ff=rand()%n+1,ff1=rand()%n+1;
	int kk=rand()%n+1;
	while(kk==ff||kk==ff1) kk=rand()%n+1;
	rep(i,1,n) {
		if(i==ff) {
			ff=0;
			cout<<100000000<<' '<<1<<" ";
		}else if(i==ff1) {
			ff1=0;
			cout<<100000000<<' '<<2<<" ";
		}else cout<<100000000<<' '<<3<<" ";
	} 
	rep(i,1,m-1) {
		cout<<1<<' '<<kk<<' '<<1<<endl;
	}
	cout<<2<<' '<<982429<<endl;
}
fire main() {
	srand(time(0));
	while(T--) {
		solve();
	}
	return false;
}