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;
int gi() {
  char cc = getchar();
  int cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const int inf = 1e9;
const int Inf = 1e8;
const int N = 4e5 + 5;
int head[N], n, q, m, rem, cnt = 1, bef = 1, st[N], top;
struct E {
  int to, next, w;
} e[N << 1];
struct SuffixTree {
  int ch[30], lef, len, lk;
} t[N];
char s[N];
int node(int l, int len) {
  ++cnt, t[cnt].len = len, t[cnt].lef = l, t[cnt].lk = 1;
  return cnt;
}
void insert(int x) {
  ++m, ++rem, s[x] -= 'a';
  int u = s[x], lst = 1;
  while (rem) {
    while (rem > t[t[bef].ch[(int)s[m - rem + 1]]].len)
      rem -= t[bef = t[bef].ch[(int)s[m - rem + 1]]].len;
    int &d = t[bef].ch[(int)s[m - rem + 1]], c = s[t[d].lef + rem - 1];
    if (!d || c == u) {
      t[lst].lk = bef, lst = bef;
      if (!d)
        d = node(m, inf);
      else
        break;
    } else {
      int np = node(t[d].lef, rem - 1);
      t[d].lef += (rem - 1), t[d].len -= (rem - 1);
      t[np].ch[c] = d, t[np].ch[u] = node(m, inf);
      t[lst].lk = d = np, lst = d;
    }
    (bef == 1) ? --rem : bef = t[bef].lk;
  }
}
int idx, dfn[N], dis[N], Fa[N], sz[N], son[N], Dep[N], Top[N], Id[N];
void DFS(int x, int fa, int dep) {
  sz[x] = 1, Dep[x] = Dep[fa] + 1, Fa[x] = fa;
  if (t[x].len > Inf) {
    t[x].len = n - t[x].lef + 1, dep += t[x].len, dis[x] = dep;
    Id[n - dep + 1] = x;
    return;
  }
  dep += t[x].len, dis[x] = dep;
  for (int i = (0); i <= (26); ++i) {
    int v = t[x].ch[i];
    if (!v) continue;
    DFS(v, x, dep), sz[x] += sz[v];
    if (sz[son[x]] <= sz[v]) son[x] = v;
  }
}
void DFS2(int x, int high) {
  Top[x] = high, dfn[x] = ++idx;
  if (son[x]) DFS2(son[x], high);
  for (int i = (0); i <= (26); ++i) {
    int v = t[x].ch[i];
    if ((!v) || (v == son[x])) continue;
    DFS2(v, v);
  }
}
int LCA(int x, int y) {
  while (Top[x] != Top[y]) {
    if (Dep[Top[x]] < Dep[Top[y]]) swap(x, y);
    x = Fa[Top[x]];
  }
  return (Dep[x] < Dep[y]) ? x : y;
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
void add(int x, int y, int z) { e[++cnt] = (E){y, head[x], z}, head[x] = cnt; }
void Add(int x, int y) { add(x, y, dis[y] - dis[x]); }
void ins(int x) {
  if (st[top] == 1) {
    st[++top] = x;
    return;
  }
  int lca = LCA(st[top], x);
  if (lca == x) return;
  while ((top > 1) && (dfn[st[top - 1]] > dfn[lca]))
    Add(st[top - 1], st[top]), --top;
  if (dfn[lca] < dfn[st[top]]) Add(lca, st[top]), --top;
  if (dfn[lca] > dfn[st[top]]) st[++top] = lca;
  st[++top] = x;
}
int fA[N], fB[N];
long long Ans;
void dfs(int x, int fa) {
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    dfs(v, x), fA[x] += fA[v], fB[x] += fB[v];
    head[v] = fA[v] = fB[v] = 0;
  }
  Ans += 1ll * fA[x] * fB[x] * (dis[x] - dis[fa]);
}
signed main() {
  n = gi(), q = gi(), scanf("%s", s + 1);
  s[++n] = 'z' + 1, t[0].len = inf;
  for (int i = (1); i <= (n); ++i) insert(i);
  --n, DFS(1, 0, 0), DFS2(1, 1);
  cnt = 0;
  vector<int> A, B, C;
  while (q--) {
    int l1 = gi(), l2 = gi(), lim = l1 + l2;
    A.resize(l1 + 1), B.resize(l2 + 1), cnt = 0;
    for (int i = (1); i <= (l1); ++i) A[i] = gi(), A[i] = Id[A[i]], ++fA[A[i]];
    for (int i = (1); i <= (l2); ++i) B[i] = gi(), B[i] = Id[B[i]], ++fB[B[i]];
    for (int i = (1); i <= (l1); ++i) C.push_back(A[i]);
    for (int i = (1); i <= (l2); ++i) C.push_back(B[i]);
    sort(C.begin(), C.end(), cmp);
    st[top = 1] = 1;
    for (int i = 0; i < lim; ++i) ins(C[i]);
    while (top > 1) Add(st[top - 1], st[top]), --top;
    dfs(1, 1), printf("%lld\n", Ans), Ans = 0;
    A.clear(), B.clear(), C.clear(), fA[1] = fB[1] = head[1] = 0;
  }
  return 0;
}