Problem: Let LCP(s, t) be the length of the longest common prefix of strings s and t. Also let s[x ... y] be the substring of s from index x to index y (inclusive). For example, if s =  "abcde", then s[1 ... 3] = "abc", s[2 ... 5] = "bcde".

You are given a string s of length n and q queries. Each query is a pair of integer sets a_1, a_2, ..., a_k and b_1, b_2, ..., b_l. Calculate ∑_{i = 1}^{i = k} ∑_{j = 1}^{j = l}{LCP(s[a_i ... n], s[b_j ... n])} for each query.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the length of string s and the number of queries, respectively.

The second line contains a string s consisting of lowercase Latin letters (|s| = n).

Next 3q lines contains descriptions of queries — three lines per query. The first line of each query contains two integers k_i and l_i (1 ≤ k_i, l_i ≤ n) — sizes of sets a and b respectively.

The second line of each query contains k_i integers a_1, a_2, ... a_{k_i} (1 ≤ a_1 < a_2 < ... < a_{k_i} ≤ n) — set a.

The third line of each query contains l_i integers b_1, b_2, ... b_{l_i} (1 ≤ b_1 < b_2 < ... < b_{l_i} ≤ n) — set b.

It is guaranteed that ∑_{i = 1}^{i = q}{k_i} ≤ 2 ⋅ 10^5 and ∑_{i = 1}^{i = q}{l_i} ≤ 2 ⋅ 10^5.

Output

Print q integers — answers for the queries in the same order queries are given in the input.

Example

Input

7 4
abacaba
2 2
1 2
1 2
3 1
1 2 3
7
1 7
1
1 2 3 4 5 6 7
2 2
1 5
1 5


Output

13
2
12
16

Note

Description of queries: 

  1. In the first query s[1 ... 7] = abacaba and s[2 ... 7] = bacaba are considered. The answer for the query is LCP(abacaba, abacaba) + LCP(abacaba, bacaba) + LCP(bacaba, abacaba) + LCP(bacaba, bacaba) = 7 + 0 + 0 + 6 = 13.
  2. In the second query s[1 ... 7] = abacaba, s[2 ... 7] = bacaba, s[3 ... 7] = acaba and s[7 ... 7] = a are considered. The answer for the query is LCP(abacaba, a) + LCP(bacaba, a) + LCP(acaba, a) = 1 + 0 + 1 = 2.
  3. In the third query s[1 ... 7] = abacaba are compared with all suffixes. The answer is the sum of non-zero values: LCP(abacaba, abacaba) + LCP(abacaba, acaba) + LCP(abacaba, aba) + LCP(abacaba, a) = 7 + 1 + 3 + 1 = 12.
  4. In the fourth query s[1 ... 7] = abacaba and s[5 ... 7] = aba are considered. The answer for the query is LCP(abacaba, abacaba) + LCP(abacaba, aba) + LCP(aba, abacaba) + LCP(aba, aba) = 7 + 3 + 3 + 3 = 16. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int np = 4e5 + 5;
template <typename _T>
inline void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s < '0' || s > '9') {
    f = 1;
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + s - '0';
    s = getchar();
  }
  x *= f;
}
int fa[np << 1], dep[np << 1], siz[np << 1], son[np << 1];
int lx[np << 1], reg[np << 1], id, top[np << 1], qaq[np << 1];
int len_;
struct Node {
  int l, r;
  long long sum, tag, k;
  Node *ls, *rs;
  inline bool inrange(int L, int R) { return L <= l && r <= R; }
  inline bool outofrange(int L, int R) { return r < L || R < l; }
  inline void pushup() { sum = ls->sum + rs->sum; }
  inline void maketag(int vl) {
    sum += k * vl;
    tag += vl;
  }
  inline void pushdown() {
    if (tag) {
      ls->maketag(tag);
      rs->maketag(tag);
      tag = 0;
    }
  }
  inline void upd(int L, int R, int vl) {
    if (inrange(L, R)) {
      maketag(vl);
    } else {
      if (!outofrange(L, R)) {
        pushdown();
        ls->upd(L, R, vl);
        rs->upd(L, R, vl);
        pushup();
      }
    }
  }
  inline long long query(int L, int R) {
    if (inrange(L, R)) {
      return sum;
    } else {
      if (!outofrange(L, R)) {
        pushdown();
        return ls->query(L, R) + rs->query(L, R);
      } else
        return 0;
    }
  }
  inline void dfs() {
    int kop = 0;
    int saio = 0;
    kop = saio * 10 + 1;
    if (ls) ls->dfs();
    int op = 1;
    int ans = 7;
    ans = op * ans;
    if (rs) rs->dfs();
    return;
  }
} mem[np << 2], *pool = mem, *rot;
int qlen[np << 1];
inline Node *New() { return ++pool; }
inline Node *build(int L, int R) {
  Node *u = New();
  u->l = L, u->r = R;
  if (L == R) {
    u->ls = u->rs = NULL;
    u->sum = 0;
    u->k = qlen[reg[L]];
  } else {
    int mid = L + R >> 1;
    u->ls = build(L, mid);
    u->rs = build(mid + 1, R);
    u->k = (u->ls->k) + (u->rs->k);
  }
  return u;
}
int head[np << 1], ver[np << 1], nxt[np << 1], tit;
int a[np << 1], b[np << 1];
inline void add(int x, int y) {
  ver[++tit] = y;
  nxt[tit] = head[x];
  head[x] = tit;
}
struct SAM {
  int link[np << 1], len[np << 1];
  int son[np << 1][27];
  int cnt, la;
  inline void add(int x) {
    int p, k, now, q;
    p = la, now = la = ++cnt, len[now] = len[p] + 1;
    qaq[len_ - (len[now] - 1)] = now;
    for (; p && !son[p][x]; son[p][x] = now, p = link[p])
      ;
    if (!p) return (void)(link[now] = 1);
    if (len[p] + 1 == len[q = son[p][x]]) return (void)(link[now] = q);
    len[k = ++cnt] = len[p] + 1, link[k] = link[q], link[q] = link[now] = k,
            memcpy(son[k], son[q], sizeof(son[q]));
    for (; p && !(son[p][x] ^ q); son[p][x] = k, p = link[p])
      ;
  }
  inline void calc_() {
    for (int i = 1; i <= cnt; i++) qlen[i] = len[i] - len[link[i]];
  }
} T;
inline void dfs_(int x, int ff) {
  fa[x] = ff;
  dep[x] = dep[ff] + 1;
  siz[x] = 1;
  for (int i = head[x], v; i; i = nxt[i]) {
    v = ver[i];
    if (v == ff) continue;
    dfs_(v, x);
    siz[x] += siz[v];
    if (siz[v] > siz[son[x]]) son[x] = v;
  }
}
inline void bfs(int x) {
  if (son[x]) {
    lx[son[x]] = ++id;
    reg[id] = son[x];
    top[son[x]] = top[x];
    bfs(son[x]);
  }
  for (int i = head[x], v; i; i = nxt[i]) {
    v = ver[i];
    if (v == fa[x]) continue;
    if (v == son[x]) continue;
    lx[v] = ++id;
    reg[id] = v;
    top[v] = v;
    bfs(v);
  }
}
inline void lca_upd(int x, int vl) {
  int fx = top[x];
  while (1 != fx) {
    rot->upd(lx[fx], lx[x], vl);
    x = fa[fx], fx = top[x];
  }
  rot->upd(1, lx[x], vl);
}
inline long long lca_query(int x) {
  long long fx = top[x], Ans = 0;
  while (1 != fx) {
    Ans += rot->query(lx[fx], lx[x]);
    x = fa[fx], fx = top[x];
  }
  Ans += rot->query(1, lx[x]);
  return Ans;
}
char c[np << 1];
signed main() {
  int q;
  read(len_);
  read(q);
  scanf("%s", c + 1);
  T.cnt = T.la = 1;
  for (int i = len_; i >= 1; i--) T.add(c[i] - 'a' + 1);
  T.calc_();
  for (int i = 1; i <= T.cnt; i++) add(i, T.link[i]), add(T.link[i], i);
  dep[0] = 0;
  dfs_(1, 0);
  lx[1] = ++id;
  top[1] = 1;
  reg[id] = 1;
  bfs(1);
  rot = build(1, T.cnt);
  for (int i = 1, a_, b_; i <= q; i++) {
    read(a_);
    read(b_);
    long long Ans = 0;
    for (int i = 1; i <= a_; i++) read(a[i]), a[i] = qaq[a[i]];
    for (int i = 1; i <= b_; i++) read(b[i]), b[i] = qaq[b[i]];
    for (int i = 1; i <= b_; i++) lca_upd(b[i], 1);
    for (int i = 1; i <= a_; i++) Ans += lca_query(a[i]);
    for (int i = 1; i <= b_; i++) lca_upd(b[i], -1);
    printf("%lld\n", Ans);
  }
}