#include<bits/stdc++.h>
#define int long long
#define N 2000005
using namespace std;

int n, m, B, root;
vector<int>p[N];
int dep[N], ans[N], dfn[N], len, tot;

struct RMQ_LCA { //O(n log n) - O(1) LCA
    int z[N], r[N], lg[N], tot;
    int f[N][17];
    void dfs(int x, int fa) {
        dep[x] = dep[fa] + 1;
        r[++tot] = x;
        dfn[x] = ++len;
        z[x] = tot;
        for(auto y : p[x]) {
            dfs(y, x);
            r[++tot] = x;
        }
    }
    int Min(int x, int y) {
        return dep[x] < dep[y] ? x : y;
    }
    void init() {
        dfs(root, 0);
        for(int i = 2; i <= tot; i++) lg[i] = lg[i / 2] + 1;
        for(int i = 1; i <= tot; i++) f[i][0] = r[i];
        for(int j = 1; j <= 15; j++) 
        for(int i = 1; i + (1 << j) - 1 <= tot; i++) f[i][j] = Min(f[i][j - 1], f[i + (1 << (j - 1)) - 1][j - 1]);
    }
    int LCA(int x, int y) {
        x = z[x], y = z[y];
        if(x > y) swap(x, y);
        int h = lg[y - x + 1];
        return Min(f[x][h], f[y - (1 << h) + 1][h]);
    }
}ljm;

int tim1;
struct DS { //分块：O(1)修改, O(sqrt(n))查询
    int v1[N], v2[N];
    int X[N], Y[N];
    int Get(int x) {
        return ceil((double)x * 1.0 / B);
    }
    void update(int x, int y) {
        X[++tim1] = x; Y[++tim1] = y;
        v1[Get(x)] += y;
        v2[x] += y;
    }
    int query(int L, int R) {
        int res = 0;
        for(int i = L; i <= R; ) {
            if(i % B == 1 && i + B - 1 <= R) res += v1[Get(i)], i += B;
            else res += v2[i], i++;
        }
        return res;
    }
    void Go(int x) { //撤销回到时间轴为x时
        while(tim1 > x) {
            v1[Get(X[tim1])] -= Y[tim1];
            v1[X[tim1]] -= Y[tim1];
            tim1--;
        }
    }
}ds;

struct List {
    int Pre, Nxt;
}t[N];

struct node {
    int l, r, x, id;
}z[N], h[N];
bool cmp1(node A, node B) {
    return A.l < B.l;
}
bool cmp2(node A, node B) {
    return A.r > B.r;
}

int v[N];
bool cmp(int x, int y) {
    return dfn[x] < dfn[y];
}
vector<int>r[N];
void Prepare() {
    for(int i = 1; i <= n; i++) v[i] = i;
    sort(v + 1, v + n + 1, cmp);
    for(int i = 1; i <= n; i++) r[dep[i]].push_back(v[i]);
    for(int d = 1; d <= n; d++) {
        for(int i = 0; i < r[d].size(); i++) {
            if(i == 0) t[r[d][i]].Pre = -1;
            else t[r[d][i]].Pre = r[d][i - 1];
            if(i == r[d].size() - 1) t[r[d][i]].Nxt = -1;
            else {
                t[r[d][i]].Nxt = r[d][i + 1];
                ds.update(d - dep[ljm.LCA(r[d][i], r[d][i + 1])], 1);
            }
        }
    }
}

int tim2;
int X[N], Y[N], opt[N];
void Go(int x) {
    while(tim2 > x) {
        if(opt[tim2] == 1) t[X[tim2]].Pre = Y[tim2];
        else t[X[tim2]].Nxt = Y[tim2];
        tim2--;
    }
}
void Del(int x) {
    if(t[x].Pre != -1) {
        ds.update(dep[x] - dep[ljm.LCA(x, t[x].Pre)], -1);
        tim2++; X[tim2] = t[x].Pre, Y[tim2] = t[t[x].Pre].Nxt, opt[tim2] = 2;
        t[t[x].Pre].Nxt = t[x].Nxt; 
    }
    if(t[x].Nxt != -1) {
        ds.update(dep[x] - dep[ljm.LCA(x, t[x].Nxt)], -1);
        tim2++; X[tim2] = t[x].Nxt, Y[tim2] = t[t[x].Nxt].Pre, opt[tim2] = 1;
        t[t[x].Nxt].Pre = t[x].Pre; 
    }
    if(t[x].Pre != -1 && t[x].Nxt != -1) 
        ds.update(dep[x] - dep[ljm.LCA(t[x].Pre, t[x].Nxt)], 1);
}

void Sol(int L, int R) { //处理左端点在[L,R]的询问
    sort(h + 1, h + tot + 1, cmp2);
    for(int i = 1; i < L; i++) Del(i);
    int curL = L, curR = n;
    for(int i = 1; i <= tot; i++) {
        while(curR > h[i].r) Del(curR--);
        int lst1 = tim1, lst2 = tim2;
        while(curL < h[i].l) Del(curL++);
        ans[h[i].id] = ds.query(h[i].x + 1, n);
        ds.Go(lst1); Go(lst2);
        curL = L;
    }
    ds.Go(0); Go(0);
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    cin >> n >> m; B = sqrt(n);
    for(int i = 1, fa; i <= n; i++) {
        cin >> fa;
        if(fa) p[fa].push_back(i);
        else root = i;
    }
    ljm.init(); Prepare();
    for(int i = 1; i <= m; i++) cin >> z[i].l >> z[i].r >> z[i].x;
    sort(z + 1, z + m + 1, cmp1);
    int j = 0;
    for(int i = 1; i <= n; i += B) {
        tot = 0;
        while(j + 1 <= m && z[j + 1].l <= i + B - 1) j++, h[++tot] = z[j];
        Sol(i, min(n, i + B - 1)); 
    }
    for(int i = 1; i <= m; i++) cout << ans[i] << endl;
    return 0;
}
/*
7 5
3 1 0 5 3 5 1
1 3 1
5 7 2
1 5 1
4 7 1
4 7 2
*/


//start : 2025/1/24 19:03