Conan and Agasa play a Card Game

Conan and Agasa play a Card Game

Edogawa Conan got tired of solving cases, and invited his friend, Professor Agasa, over. They decided to play a game of cards. Conan has n cards, and the i-th card has a number ai written on it.

They take turns playing, starting with Conan. In each turn, the player chooses a card and removes it. Also, he removes all cards having a number strictly lesser than the number on the chosen card. Formally, if the player chooses the i-th card, he removes that card and removes the j-th card for all j such that aj < ai.

A player loses if he cannot make a move on his turn, that is, he loses if there are no cards left. Predict the outcome of the game, assuming both players play optimally.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of cards Conan has.

The next line contains n integers a1, a2, …, an (1 ≤ ai ≤ 105), where ai is the number on the i-th card.

Output

If Conan wins, print “Conan” (without quotes), otherwise print “Agasa” (without quotes).

Examples

nput
3
4 5 7
Output
Conan
Input
2
1 1
Output
Agasa

Note

In the first example, Conan can just choose the card having number 7 on it and hence remove all the cards. After that, there are no cards left on Agasa’s turn.

In the second example, no matter which card Conan chooses, there will be one one card left, which Agasa can choose. After that, there are no cards left when it becomes Conan’s turn again.

题目大意：

现有n张卡片，每张卡片上都有一个数字，柯南首先取出一张卡片，并且将所有数字小于这张卡片的数字的卡片取走。然后换下一个玩家操作，谁没有卡片可取谁就输了。

思路：

造样例观察可发现，如果某种卡片出现次数为奇数，那么必定是先手赢，如果所有卡片出现次数均为偶数，则后手赢。
代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[101000];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        int temp;
        cin>>temp;
        a[temp]++;
    }
    for(int i=0;i<=100000;i++)
        if(a[i]%2)
        {
         cout<<"Conan"<<endl;
         return 0;
        }
    cout<<"Agasa"<<endl;
    return 0;
}