Werewolf（狼人杀） is a game in which the players are partitioned into two parties: the werewolves and the human beings. Suppose that in a game,

• player #1 said: "Player #2 is a werewolf.";
• player #2 said: "Player #3 is a human.";
• player #3 said: "Player #4 is a werewolf.";
• player #4 said: "Player #5 is a human."; and
• player #5 said: "Player #4 is a human.".

Given that there were 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. Can you point out the werewolves?

Now you are asked to solve a harder version of this problem: given that there were N players, with 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. You are supposed to point out the werewolves.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (5≤N≤100). Then N lines follow and the i-th line gives the statement of the i-th player (1≤i≤N), which is represented by the index of the player with a positive sign for a human and a negative sign for a werewolf.

Output Specification:

If a solution exists, print in a line in ascending order the indices of the two werewolves. The numbers must be separated by exactly one space with no extra spaces at the beginning or the end of the line. If there are more than one solution, you must output the smallest solution sequence -- that is, for two sequences A=a[1],...,a[M] and B=b[1],...,b[M], if there exists 0≤kNo Solution.

Sample Input 1:

5
-2
+3
-4
+5
+4

Sample Output 1:

1 4

Sample Input 2:

6
+6
+3
+1
-5
-2
+4

Sample Output 2 (the solution is not unique):

1 5

Sample Input 3:

5
-2
-3
-4
-5
-1

Sample Output 3:

No Solution

两个狼人两个骗子，其中有一个狼人是骗子，四层for循环来枚举：

两个for循环枚举两个狼人，这样答案输出时可以保证是最小序列，然后分别考虑两个狼人各自撒谎的时候，第三层for循环枚举撒谎的人类，然后更改两个骗子的陈述，取相反数即可，第四层for循环验证这样的假设情况下所有人的陈述是否矛盾，如果不矛盾直接输出两个狼人，return 0；否则继续下一次枚举，如果所有枚举的情况都有矛盾，输出No Solution。

#include 
#include 
#include 
using namespace std;
int n;
int a[110]; //输入、真正陈述
bool flag;int main() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i];}for (int i = 1; i <= n; i++) {for (int j = i + 1; j <= n; j++) { //枚举狼人a[i] = -a[i]; //第一个狼人说谎for (int k = 1; k <= n; k++) { //枚举说谎的人类flag = 0;if (k != i && k != j) {a[k] = -a[k];for (int c = 1; c <= n; c++) { //验证说法是否矛盾if (a[c] < 0 && abs(a[c]) != i && abs(a[c]) != j) {flag = 1;break;}if (a[c] > 0 && (a[c] == i || a[c] == j)) {flag = 1;break;}}if (!flag) {cout << i << ' ' << j;return 0;}a[k] = -a[k];}}a[i] = -a[i];a[j] = -a[j]; //第二个狼人说谎for (int k = 1; k <= n; k++) { //枚举说谎的人类flag = 0;if (k != i && k != j) {a[k] = -a[k];for (int c = 1; c <= n; c++) { //验证说法是否矛盾if (a[c] < 0 && abs(a[c]) != i && abs(a[c]) != j) {flag = 1;break;}if (a[c] > 0 && (a[c] == i || a[c] == j)) {flag = 1;break;}}if (!flag) {cout << i << ' ' << j;return 0;}a[k] = -a[k];}}a[j] = -a[j];}}cout << "No Solution";return 0;
}