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 MAXN = 400005;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
char s[MAXN];
int n, Q, pos[MAXN];
struct SuffixAutomaton {
  int tot, last, ch[MAXN][26], fa[MAXN], maxl[MAXN];
  int Extend(int x) {
    int p = last, np = ++tot;
    maxl[last = np] = maxl[p] + 1;
    for (; p && !ch[p][x]; p = fa[p]) {
      ch[p][x] = np;
    }
    if (!p) {
      fa[np] = 1;
    } else {
      int q = ch[p][x];
      if (maxl[q] == maxl[p] + 1) {
        fa[np] = q;
      } else {
        int nq = ++tot;
        maxl[nq] = maxl[p] + 1;
        memcpy(ch[nq], ch[q], sizeof(ch[nq]));
        fa[nq] = fa[q];
        fa[q] = fa[np] = nq;
        for (; ch[p][x] == q; p = fa[p]) {
          ch[p][x] = nq;
        }
      }
    }
    return last;
  }
  void build() {
    tot = last = 1;
    for (int i = n; i >= 1; --i) {
      pos[i] = Extend(s[i] - 'a');
    }
  }
} sam;
struct Edge {
  int v, nxt;
} e[MAXN];
int first[MAXN], eCnt;
inline void AddEdge(int u, int v) {
  e[++eCnt].v = v;
  e[eCnt].nxt = first[u];
  first[u] = eCnt;
}
int fa[MAXN], dep[MAXN], siz[MAXN], son[MAXN], top[MAXN], dfn[MAXN], clk;
int na, nb, tot, vir[MAXN], stk[MAXN], tp;
int isa[MAXN], isb[MAXN], asiz[MAXN], bsiz[MAXN];
long long ans;
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
void DFS1(int u) {
  siz[u] = 1;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa[u]) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    DFS1(v);
    if (siz[v] > siz[son[u]]) son[u] = v;
    siz[u] += siz[v];
  }
}
void DFS2(int u, int topf) {
  top[u] = topf;
  dfn[u] = ++clk;
  if (son[u]) DFS2(son[u], topf);
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa[u] || v == son[u]) continue;
    DFS2(v, v);
  }
}
inline int LCA(int u, int v) {
  while (top[u] ^ top[v]) {
    (dep[top[u]] > dep[top[v]]) ? u = fa[top[u]] : v = fa[top[v]];
  }
  return (dep[u] < dep[v]) ? u : v;
}
void Build() {
  tot = sam.tot;
  for (int i = 2; i <= tot; ++i) {
    AddEdge(sam.fa[i], i);
  }
  DFS1(1);
  DFS2(1, 1);
  memset(first, 0, sizeof(first));
  eCnt = 0;
}
void GetAns(int u) {
  asiz[u] = isa[u];
  bsiz[u] = isb[u];
  long long tmp = 0;
  if (isa[u] && isb[u]) ++tmp;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    GetAns(v);
    tmp += 1LL * asiz[u] * bsiz[v] + 1LL * bsiz[u] * asiz[v];
    asiz[u] += asiz[v];
    bsiz[u] += bsiz[v];
  }
  ans += tmp * sam.maxl[u];
}
void Clean(int u) {
  isa[u] = isb[u] = asiz[u] = bsiz[u] = 0;
  for (int i = first[u]; i; i = e[i].nxt) {
    Clean(e[i].v);
  }
  first[u] = 0;
}
void Work() {
  read(na);
  read(nb);
  tot = 0;
  for (int i = 1; i <= na; ++i) {
    int x;
    read(x);
    vir[++tot] = pos[x];
    isa[pos[x]] = 1;
  }
  for (int i = 1; i <= nb; ++i) {
    int x;
    read(x);
    vir[++tot] = pos[x];
    isb[pos[x]] = 1;
  }
  sort(vir + 1, vir + tot + 1, cmp);
  tot = unique(vir + 1, vir + tot + 1) - (vir + 1);
  stk[tp = 1] = vir[1];
  for (int i = 2; i <= tot; ++i) {
    int lca = LCA(vir[i], stk[tp]);
    while (tp && dep[lca] < dep[stk[tp]]) {
      if (dep[lca] >= dep[stk[tp - 1]]) {
        AddEdge(lca, stk[tp]);
        if (stk[--tp] != lca) stk[++tp] = lca;
        break;
      }
      --tp;
      AddEdge(stk[tp], stk[tp + 1]);
    }
    stk[++tp] = vir[i];
  }
  while (--tp) {
    AddEdge(stk[tp], stk[tp + 1]);
  }
  ans = 0;
  GetAns(stk[1]);
  printf("%lld\n", ans);
  Clean(stk[1]);
  eCnt = 0;
}
void solve() {
  read(n);
  read(Q);
  scanf("%s", s + 1);
  sam.build();
  Build();
  while (Q--) {
    Work();
  }
}
int main() {
  solve();
  return 0;
}