題目連結：http://codeforces.com/problemset/problem/832/D

Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other.

The boys decided to have fun and came up with a plan. Namely, in some day in the morning Misha will ride the underground from station s

 to station f by the shortest path, and will draw with aerosol an ugly text "Misha was here" on every station he will pass through (including sand f). After that on the same day at evening Grisha will ride from station t to station f by the shortest path and will count stations with Misha's text. After that at night the underground workers will wash the texts out, because the underground should be clean.

The boys have already chosen three stations a, b and c for each of several following days, one of them should be station s on that day, another should be station f, and the remaining should be station t. They became interested how they should choose these stations s

, f, t so that the number Grisha will count is as large as possible. They asked you for help.

Input

The first line contains two integers n and q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of stations and the number of days.

The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi ≤ n). The integer pi means that there is a route between stations pi and i. It is guaranteed that it's possible to reach every station from any other.

The next q lines contains three integers a, b and c each (1 ≤ a, b, c ≤ n) — the ids of stations chosen by boys for some day. Note that some of these ids could be same.

Output

Print q lines. In the i-th of these lines print the maximum possible number Grisha can get counting when the stations s, t and f are chosen optimally from the three stations on the i-th day.

Examples

input

3 2
1 1
1 2 3
2 3 3

output

2
3

input

4 1
1 2 3
1 2 3

output

2

Note

In the first example on the first day if s = 1, f = 2, t = 3, Misha would go on the route 1  2, and Grisha would go on the route 3  1  2. He would see the text at the stations 1 and 2. On the second day, if s = 3, f = 2, t = 3, both boys would go on the route 3  1  2. Grisha would see the text at 3 stations.

In the second examle if s = 1, f = 3, t = 2, Misha would go on the route 1  2  3, and Grisha would go on the route 2  3 and would see the text at both stations.

題目大意：給出n個車站，m次訪問

n-1個數字，意思為i-pi的道路是通路（雙向）

之後是m次訪問，輸入三個點，任意一個是終點，另兩個是起點，找出重合點最多的路線，輸出重合點的數量

其實並不難，關鍵是如何確定重合最多的點，因為就三個點，可以直接處理處三個情況，然後取最大的即可；

對於a，b，c三點；很容易從圖中看到，如果以a，b為起點，那麼(lca(a,c)+lca(b,c)-lca(a,b))/2就是他們的重合的邊了，那麼因為有三條邊，直接找出三條邊的長度然後取最長的那個即可：

ac：

//#pragma comment(linker, "/STACK:1024000000,1024000000") 

#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
//#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  

#define ll long long  
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int MAXN=101000;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;

struct node{
	int v,w,nxt;
	node(int _v=0,int _nxt=0):
	v(_v),nxt(_nxt){}
}edge[MAXN<<1];
int head[MAXN],ecnt;
int fa[MAXN][30],deep[MAXN];
int len[MAXN][30];
int n,m;
void intt()
{
	clean(len,0);
	clean(head,-1);
	clean(deep,0);
	clean(fa,-1);
	ecnt=0;
}
 
void add(int u,int v)
{
	edge[ecnt]=node(v,head[u]);
	head[u]=ecnt++;
}
/*---------------板子-----------------*/
void dfs(int u)
{
	for(int i=head[u];i+1;i=edge[i].nxt)
	{
		int temp=edge[i].v;
		if(deep[temp]==0)
		{
			deep[temp]=deep[u]+1;
			fa[temp][0]=u;
			len[temp][0]=1;
			int up=0,pre=u;
			while(fa[pre][up]>=0)
			{
				fa[temp][up+1]=fa[pre][up];
				len[temp][up+1]=len[temp][up]+len[pre][up];
				pre=fa[pre][up++];
			}
			dfs(temp);
		}
	}
}

int lca(int a,int b)
{
	int ans=0;
	if(deep[a]<deep[b])
		swap(a,b);
	int lim=log2(deep[a])+1;
	for(int i=lim;i>=0;--i)
	{
		if(deep[fa[a][i]]>=deep[b])
		{
			ans+=len[a][i];
			a=fa[a][i];
		}
	}
	if(a==b)
		return ans;
	for(int i=lim;i>=0;--i)
	{
		if(fa[a][i]!=fa[b][i])
		{
			ans=ans+len[a][i];
			a=fa[a][i];
			ans=ans+len[b][i];
			b=fa[b][i];
		}
	}
	ans=ans+len[a][0]+len[b][0];
	return ans;
}

int main()
{
	std::ios::sync_with_stdio(false);
	intt();
	cin>>n>>m;
	int num;
	for(int i=2;i<=n;++i)
	{
		cin>>num;
		add(i,num);
		add(num,i);
	}
	deep[1]=1;
	dfs(1);
	int a,b,c;
	for(int i=1;i<=m;++i)
	{
		cin>>a>>b>>c;
		//以任意點為終點，另兩個點為起點，重合的最多的點 
		int lab=lca(a,b);
		int lac=lca(a,c);
		int lbc=lca(b,c);
		int ans1=(lac+lbc-lab)/2+1;
		int ans2=(lac+lab-lbc)/2+1;
		int ans3=(lab+lbc-lac)/2+1;
		cout<<max(ans1,max(ans2,ans3))<<endl;
		
	}
}