C. Game On Leaves

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这是一个关于在无根树上进行游戏的问题，两个玩家Ayush和Ashish轮流选择一个叶节点和连接它的边移除。叶节点是度数小于或等于1的节点。特殊节点xx的移除者获胜。Ayush先手，需要确定在最优情况下谁会赢得游戏。输入包含树的节点数和特殊节点，以及树的边信息。输出是获胜者的名字。示例展示了不同情况下的获胜者。

链接https://codeforces.ml/contest/1363/problem/C

Ayush and Ashish play a game on an unrooted tree consisting of nn nodes numbered 11 to nn. Players make the following move in turns:

Select any leaf node in the tree and remove it together with any edge which has this node as one of its endpoints. A leaf node is a node with degree less than or equal to 11.

A tree is a connected undirected graph without cycles.

There is a special node numbered xx. The player who removes this node wins the game.

Ayush moves first. Determine the winner of the game if each player plays optimally.

Input

The first line of the input contains a single integer tt (1≤t≤10)(1≤t≤10) — the number of testcases. The description of the test cases follows.

The first line of each testcase contains two integers nn and xx (1≤n≤1000,1≤x≤n)(1≤n≤1000,1≤x≤n) — the number of nodes in the tree and the special node respectively.

Each of the next n−1n−1 lines contain two integers uu, vv (1≤u,v≤n, u≠v)(1≤u,v≤n, u≠v), meaning that there is an edge between nodes uu and vv in the tree.

Output

For every test case, if Ayush wins the game, print "Ayush", otherwise print "Ashish" (without quotes).

Examples

input

Copy

output

Copy

Ashish

input

Copy

output

Copy

Ayush

Note

For the 11st test case, Ayush can only remove node 22 or 33, after which node 11 becomes a leaf node and Ashish can remove it in his turn.

For the 22nd test case, Ayush can remove node 22 in the first move itself.

代码：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define lb double
#define INF 0x3f3f3f3f
#define maxn 200010
#define yes cout<<"Yes"<<endl
#define no cout<<"No"<<endl
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define gep(i,x,y) for(int i=x;i>=y;i--)
ll n,t,x,k,max1,s,ans,mod=1e9+7;
ll a,b,c[20001];
vector<ll>p[20001];
void dfs(int i)
{
	for(int j=0;j<p[i].size();j++)
	{
		if(c[p[i][j]]==0)
		{
			c[p[i][j]]=1;
			k=p[p[i][j]].size();
			s+=k-1;
			dfs(p[i][j]);
		}
	}
}
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>n>>x;
		for(int i=1;i<=n;i++)
		{
			c[i]=0;
			p[i].clear();
		}
		for(int i=1;i<=n-1;i++)
		{
			cin>>a>>b;
			p[a].push_back(b);
			p[b].push_back(a);
		}
		s=p[x].size();
		if(s==1||n==1)
		{
			cout<<"Ayush"<<endl;
		}
		else
		{
			c[x]=1;
			dfs(x);
			if(s%2==1)
			cout<<"Ayush"<<endl;
			else
			cout<<"Ashish"<<endl;
		}
		
	}
    
}