我奇特的脑回路的做法就是
树链剖分 + 树状数组

树状数组是那种 区间修改，区间求和，还有回溯的

当我看到别人写的是lca，直接讨论时，感觉自己的智商收到了碾压。。。

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

#define MS(x,y) memset(x,y,sizeof(x))

#define MP(x, y) make_pair(x, y)

const int INF = 0x3f3f3f3f;

int n, q;

struct Node{

    int to,next;

}edge[N << 1];

int tot;

int head[N];

void addedge(int x,int y){

    edge[tot].to = y; edge[tot].next = head[x]; head[x] = tot ++;

}

int top[N],fa[N],son[N],deep[N],num[N],p[N],fp[N],pos;

void dfs1(int x,int pre,int dep){

    deep[x] = dep;

    fa[x] = pre;

    num[x] = 1;

    for(int i = head[x]; i != -1; i = edge[i].next){

        int y = edge[i].to;

        if(y == pre) continue;

        dfs1(y,x,dep + 1);

        num[x] += num[y];

        if(son[x] == -1 || num[y] > num[son[x]])

            son[x] = y;

    }

}

void dfs2(int x,int tp){

    top[x] = tp;

    p[x] = pos++;

    fp[p[x]] = x;

    if(son[x] == -1) return;

    dfs2(son[x],tp);

    for(int i = head[x] ; i != -1; i = edge[i].next){

        int y = edge[i].to;

        if(y != son[x] && y != fa[x])

            dfs2(y,y);

    }

}

ll tree1[N]; ll tree2[N];

void Add(ll tree[], int pos, int val) {

    for(int i = pos; i <= n; i += i&-i) {

        tree[i] += val;

    }

}

ll Sum(ll tree[], int pos) {

    if(pos == 0) return 0;

    ll ans = 0;

    for(int i = pos; i; i -= i&-i) {

        ans += tree[i];

    }

    return ans;

}

vector<pair<int, int> > Resume;

void Find(int x, int y) {

    int fx = top[x]; int fy = top[y];

    while(fx != fy) {

        if(deep[fx] < deep[fy]) {

            swap(fx, fy);

            swap(x, y);

        }

        Add(tree1, p[fx], 1);

        Add(tree1, p[x]+1, -1);

        Add(tree2, p[fx], p[fx]);

        Add(tree2, p[x]+1, -p[x]-1);

        Resume.push_back(MP(p[fx], -1));

        Resume.push_back(MP(p[x]+1, 1));

        Resume.push_back(MP(-p[fx], -p[fx]));

        Resume.push_back(MP(-p[x]-1, p[x]+1));

        x = fa[fx];

        fx = top[x];

    }

    if(deep[x] > deep[y]) swap(x, y);

    Add(tree1, p[x], 1);

    Add(tree1, p[y]+1, -1);

    Add(tree2, p[x], p[x]);

    Add(tree2, p[y]+1, -p[y]-1);

    Resume.push_back(MP(p[x], -1));

    Resume.push_back(MP(p[y]+1, 1));

    Resume.push_back(MP(-p[x], -p[x]));

    Resume.push_back(MP(-p[y]-1, p[y]+1));

}

ll Total(int x, int y) {

    ll sum = 0;

    int fx = top[x]; int fy = top[y];

    while(fx != fy) {

        if(deep[fx] < deep[fy]) {

            swap(fx, fy);

            swap(x, y);

        }

        sum += 1ll*(p[x]+1)*Sum(tree1, p[x]) - Sum(tree2, p[x]) - 1ll*(p[fx])*Sum(tree1, p[fx]-1) + Sum(tree2, p[fx]-1);

        x = fa[fx];

        fx = top[x];

    }

    if(deep[x] > deep[y]) swap(x, y);

    sum += 1ll*(p[y]+1)*Sum(tree1, p[y]) - Sum(tree2, p[y]) - 1ll*(p[x])*Sum(tree1, p[x]-1) + Sum(tree2, p[x]-1);

    return sum;

}

ll solve(int a,int b, int c,int d) {

    Resume.clear();

    Find(a, b);

//  for(int i = 1; i <= n; ++i) printf("%d ", Sum(i)); printf("\n");

    ll tt = Total(c, d);

//  printf("hh %d %d %d %d %d\n",a,b,c,d, tt);

    for(int i = 0; i < Resume.size(); ++i) {

        if(Resume[i].first > 0) Add(tree1, Resume[i].first, Resume[i].second);

        else                    Add(tree2, -Resume[i].first, Resume[i].second);

    }

    return tt;

}

int main() {

    while(~scanf("%d %d", &n, &q)) {

        MS(tree1, 0); MS(tree2, 0);

        memset(head, -1, sizeof(head));

        memset(son, -1, sizeof(son));

        tot = 0; pos = 1;

        for(int i = 2; i <= n; ++i) {

            int a; scanf("%d", &a);

            addedge(a, i); addedge(i, a);

        }

        dfs1(1, 1, 1);

        dfs2(1, 1);

//      for(int i = 1; i <= n; ++i) printf("%d ", p[i]); printf("\n");

        for(int i = 0; i < q; ++i) {

            int a, b, c; scanf("%d %d %d", &a, &b, &c);

            ll ans = max(max(solve(a,b, a,c), solve(a,b, b,c)), solve(a,c, b,c));

            printf("%lld\n", ans);

        }

    }

    return 0;

}