Codeforces Round #425 (Div. 2) D. Misha, Grisha and Underground LCA

D. Misha, Grisha and Underground

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 s and 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 21 11 2 32 3 3

output

23

input

4 11 2 31 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.

一棵树，给三个节点，求它们间一个点到其余两个点路径的公共路径上经过节点数的最大值

LCA.

把两两之间的LCA求出来，设起点为A，终点为B、C，公共路径的距离即Vab+Vac-Vbc，两点之间的距离可由LCA求出。由于需要求的是节点数量，最终的结果需要+1.

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <bitset>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;const int maxn=100005,inf=0x3f3f3f3f;const ll llinf=0x3f3f3f3f3f3f3f3f; const ld pi=acos(-1.0L);int head[maxn],a[maxn*2],mind[maxn*2][20],f[maxn],d[maxn],dep[maxn*2];bool visit[maxn];int dfn,num;struct Edge {int from,to,pre,dist;};Edge edge[maxn*2];void addedge(int from,int to) {edge[num]=(Edge){from,to,head[from]};head[from]=num++;edge[num]=(Edge){to,from,head[to]};head[to]=num++;}void dfs(int now,int step) {visit[now]=1;a[++dfn]=now;f[now]=dfn;dep[dfn]=step;for (int i=head[now];i!=-1;i=edge[i].pre) {int to=edge[i].to;if (!visit[to]) {d[to]=d[now]+1;dfs(to,step+1);a[++dfn]=now;dep[dfn]=step;}}}void init(int n) {int i,j;for (i=1;i<=n;i++) mind[i][0]=i;for (j=1;(1<<j)<=n;j++) {for (i=1;i+(1<<j)-1<=n;i++) {int p=mind[i][j-1],q=mind[i+(1<<(j-1))][j-1];mind[i][j]=dep[p]<dep[q]?p:q;}}}int find(int l,int r) {int k=0;if (l>r) {int p=l;l=r;r=p;}while (1<<(k+1)<=(r-l+1)) k++;int p=mind[l][k],q=mind[r-(1<<k)+1][k];return dep[p]<dep[q]?a[p]:a[q];}int main() {memset(head,-1,sizeof(head));int n,q,i,j,x,y,z;scanf("%d%d",&n,&q);num=0;for (i=1;i<n;i++) {scanf("%d",&x);addedge(x,i+1);}mem0(visit);d[1]=dfn=0;dfs(1,0);init(dfn);for (i=1;i<=q;i++) {scanf("%d%d%d",&x,&y,&z);int pxy,pyz,pxz,dxy,dyz,dxz;pxz=find(f[x],f[z]);pyz=find(f[y],f[z]);pxy=find(f[x],f[y]);dxy=d[x]+d[y]-2*d[pxy];dyz=d[z]+d[y]-2*d[pyz];dxz=d[z]+d[x]-2*d[pxz];int ans=max(max(dxy+dyz-dxz,dyz+dxz-dxy),dxy+dxz-dyz);printf("%d\n",ans/2+1);}return 0;}