#include<bits/stdc++.h>
using namespace std;
const double PI=acos(-1);
#define int long long
#define double long double
#define endl '\n'
int qmi(int a, int k, int p) { // 求a^k mod p 求a逆元就是(a,p-2,p)
	int res = 1 % p;
	while (k) {
		if (k & 1) res = (__int128) res * a % p;//c++20才可以
		a = (__int128) a * a % p;//龟速乘就是把两个*的符号改成+
		k >>= 1;
	}
	return res;
}
int gcd (int a ,int b) {
	return b ? gcd ( b , a % b ) :a;
}
const int N=1e5+10;
int a[N];
const int p=998244353;
map<int,int>mp;
int dfs(int x) {

	if(x==0) return 1;
	if(mp[x]) return mp[x];
	mp[x]=dfs(x/2)+dfs(x/3);
	return mp[x];

}

signed main() {
	int  n,m,t;
	cin>>n>>m;
	cin>>t;

	while(t--) {
		int k=1,x=1,y=1;
		int a,b,c;
		cin>>a>>b>>c;
        if(a==0&&b==0) {//特判
            cout<<k<<" "<<x<<" "<<y<<endl;
            continue;
        }
        else {
            int k0=0;//最少多少个c让走向下一层
            if(a){
                k0=(n-1)/a+1;
            }
            if(b){
                k0=(n-1)/b+1;
            }
            if(a&&b){
                k0=min((n-1)/a,(n-1)/b)+1;
            }
            
		    k+=c/k0;
            c-=c/k0*k0;
        }
        
		while(c--) {
			x+=a,y+=b;
			if(x>n||y>m) {
				k++,x=1,y=1;
			}
		}
		cout<<k<<" "<<x<<" "<<y<<endl;
	}
	return 0;
}

https://ac.nowcoder.com/acm/contest/44246/C