先考慮不要求區間怎麼做，即我們要查詢:

\[\sum\limits_{i = } ^ n dis_{u, i} = \sum\limits_{i = 1} ^ n dep_u + dep_i - 2 \times dep_{lca(u, i)} \]

前面這一坨 \(\sum\limits_{i = 1} ^ n dep_u + dep_i = n \times dep_u + \sum\limits_{i = 1} ^ n dep_i\) 可以很快計算出來，後面那一坨實際上要求的就是 \(\sum\limits_{i = 1} ^ n dep_{lca(u, i)}\)，這實際上就是一道原題

#include<bits/stdc++.h>
using namespace std;
#define ls t[p].l
#define rs t[p].r
#define mid (l + r) / 2
#define rep(i, l, r) for(int i = l; i <= r; ++i)
#define Next(i, u) for(int i = h[u]; i; i = e[i].next)
typedef long long ll;
const int N = 150000 + 5;
const int M = 15000000 + 5;
struct node{
    int w, p;
    bool operator < (const node &x) const{
        return w == x.w ? p < x.p : w < x.w;
    }
}a[N];
struct edge{
    int v, next, w;
}e[N << 1];
struct tree{
    int l, r, tag; ll sum;
}t[M];
ll ans, dep[N], ton[N];
int n, m, u, v, w, l, r, x, y, Q, A, tot, cnt, num, last;
int s[N], h[N], rt[N], fa[N], dfn[N], dis[N], son[N], top[N];
int read(){
    char c; int x = 0, f = 1;
    c = getchar();
    while(c > '9' || c < '0'){ if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
void add(int u, int v, int w){
    e[++tot].v = v, e[tot].w = w, e[tot].next = h[u], h[u] = tot;
    e[++tot].v = u, e[tot].w = w, e[tot].next = h[v], h[v] = tot;
}
void dfs1(int u, int Fa){
    int Max = -1; s[u] = 1, fa[u] = Fa;
    Next(i, u){
        int v = e[i].v; if(v == Fa) continue;
        dep[v] = dep[u] + e[i].w, dfs1(v, u), s[u] += s[v];
        if(s[v] > Max) Max = s[v], son[u] = v;
    }
}
void dfs2(int u, int topf){
    top[u] = topf, dfn[u] = ++num, dis[num] = dep[u] - dep[fa[u]]; // dfs 序一定要在 dfs2 記錄
    if(son[u]) dfs2(son[u], topf);
    Next(i, u){
        int v = e[i].v; if(v == fa[u] || v == son[u]) continue;
        dfs2(v, v);
    }
}
void build(int &p, int l, int r){
    p = ++cnt; if(l == r) return;
    build(ls, l, mid), build(rs, mid + 1, r);
}
void update(int &p, int l, int r, int x, int y){
    if(p <= last) t[++cnt] = t[p], p = cnt; // 這樣可以節省大量空間，同一顆主席樹上不必新開節點
    t[p].sum += dis[y] - dis[x - 1];
    if(l >= x && r <= y){ ++t[p].tag; return;}
    if(y <= mid) update(ls, l, mid, x, y);
    else if(x > mid) update(rs, mid + 1, r, x, y);
    else update(ls, l, mid, x, mid), update(rs, mid + 1, r, mid + 1, y); // 標記永久化必須沿途直接更新答案，因為如果這個地方打上懶標記下次在下面修改時直接繼承左右兒子答案是有問題的
}
ll query(int p, int l, int r, int x, int y){
    if(!p || l >= x && r <= y) return t[p].sum;
    ll tmp = 1ll * t[p].tag * (dis[y] - dis[x - 1]);
    if(y <= mid) return tmp + query(ls, l, mid, x, y);
    else if(x > mid) return tmp + query(rs, mid + 1, r, x, y);
    else return tmp + query(ls, l, mid, x, mid) + query(rs, mid + 1, r, mid + 1, y);
}
void Modify(int x, int w){
    while(top[x] != 1) update(rt[w], 1, n, dfn[top[x]], dfn[x]), x = fa[top[x]];
    update(rt[w], 1, n, 1, dfn[x]);
}
ll Ask(int x, int w){
    ll ans = 0;
    while(top[x] != 1) ans += query(rt[w], 1, n, dfn[top[x]], dfn[x]), x = fa[top[x]];
    return ans + query(rt[w], 1, n, 1, dfn[x]);
}
int main(){
    n = read(), Q = read(), A = read();
    rep(i, 1, n) a[i].w = read(), a[i].p = i;
    sort(a + 1, a + n + 1);
    rep(i, 1, n - 1) u = read(), v = read(), w = read(), add(u, v, w);
    dfs1(1, 0), dfs2(1, 1);
    rep(i, 1, n) ton[i] = ton[i - 1] + dep[a[i].p];
    rep(i, 1, n) dis[i] += dis[i - 1];
    build(rt[0], 1, n); rep(i, 1, n) rt[i] = rt[i - 1], last = cnt, Modify(a[i].p, i); // 需要在主席樹上修改多次時只能這麼寫
    while(Q--){
        u = read(), x = read(), y = read();
        l = min((x + ans % A) % A, (y + ans % A) % A), r = max((x + ans % A) % A, (y + ans % A) % A);
        x = l, y = r;
        r = upper_bound(a + 1, a + n + 1, (node){y, n}) - a - 1;
        l = lower_bound(a + 1, a + n + 1, (node){x, 0}) - a; // 這裡不能先離散化年齡再查
        ans = 1ll * (r - l + 1) * dep[u] + ton[r] - ton[l - 1] - 2ll * (Ask(u, r) - Ask(u, l - 1)); // 不需要直接訪問區間資訊時直接字首和更方便
        printf("%lld\n", ans);
    }
    return 0;
}