Codeforces Round #266 (Div. 2) B. Wonder Room_html/css_WEB-ITnose

the start of the new academic year brought about the problem of accommodation students into dormitories. one of such dormitories has a a?×?b square meter wonder room. the caretaker wants to accommodate exactly n students there. but the law says that there must be at least 6 square meters per student in a room (that is, the room for n students must have the area of at least 6n square meters). the caretaker can enlarge any (possibly both) side of the room by an arbitrary positive integer of meters. help him change the room so as all n students could live in it and the total area of the room was as small as possible.

Input

The first line contains three space-separated integers n, a and b (1?≤?n,?a,?b?≤?109) ? the number of students and the sizes of the room.

Output

Print three integers s, a1 and b1 (a?≤?a1; b?≤?b1) ? the final area of the room and its sizes. If there are multiple optimal solutions, print any of them.

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Sample test(s)

Input

3 3 5

Output

183 6

Input

2 4 4

Output

164 4题意：安排宿舍大小，每个学生至少需要6平方米的大小，求n个学生，原本a*b大小的宿舍，再满足要求的情况下，新的宿舍由旧的宿舍扩建后的大小思路：搜索，起初10^9是过不了的，但是我们枚举一边a的大小后，对于另一边b，如果加一个条件我们起初设a<=b，那么再枚举到a>b的时候，等于说会出现对称的情况了，这样数据量就小一半了<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>typedef long long ll;using namespace std;int main() {	ll n, a, b;	cin >> n >> a >> b;	n *= 6;	int flag = 0;	if (a > b) { // 初始让x < y		swap(a, b);		flag = 1;	}	ll ans = 2000000000000000000ll;;	ll nx = -1, ny = -1;	for (ll i = 1; i <= n; i++) {		ll x = i;		ll y = (n + x - 1) / x;		if (x > y) // 不符合，即出现对称			break;		if (x < a) x = a;		if (y < b) y = b;		if (x * y < ans) {			nx = x;			ny = y;			ans = x * y;		}	}	if (flag) 		swap(nx, ny);	cout << ans << endl;	cout << nx << " " << ny << endl;	return 0;}