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>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
void umax(int& a, int t) {
  if (t > a) a = t;
}
bool umin(int& a, int t) {
  if (t < a) return a = t, 1;
  return 0;
}
struct SAM {
  int t[400011][26], pre[400011], len[400011];
  int last, tot;
  SAM() { tot = last = 1; }
  void insert(int w) {
    int pos = last, cur = ++tot;
    len[cur] = len[pos] + 1, last = cur;
    while (pos && !t[pos][w]) t[pos][w] = cur, pos = pre[pos];
    if (!pos) {
      pre[cur] = 1;
      return;
    }
    int nxt = t[pos][w];
    if (len[nxt] == len[pos] + 1)
      pre[cur] = nxt;
    else {
      int tmp = ++tot;
      len[tmp] = len[pos] + 1;
      memcpy(t[tmp], t[nxt], sizeof t[nxt]);
      pre[tmp] = pre[nxt], pre[nxt] = pre[cur] = tmp;
      while (pos && t[pos][w] == nxt) t[pos][w] = tmp, pos = pre[pos];
    }
  }
} sam;
struct Segment_Tree {
  struct node {
    long long sum, k, tag;
  } t[400011 << 2 | 1];
  void pushdown(unsigned l, unsigned r, unsigned num) {
    if (!t[num].tag) return;
    unsigned mid = (l + r) >> 1;
    t[num << 1].sum += t[num << 1].k * t[num].tag,
        t[num << 1].tag += t[num].tag;
    t[num << 1 | 1].sum += t[num << 1 | 1].k * t[num].tag,
        t[num << 1 | 1].tag += t[num].tag;
    t[num].tag = 0;
  }
  void build(int* arr, unsigned l = 1, unsigned r = sam.tot, unsigned num = 1) {
    if (l == r)
      t[num].k = arr[l];
    else {
      unsigned mid = (l + r) >> 1;
      build(arr, l, mid, num << 1), build(arr, mid + 1, r, num << 1 | 1);
      t[num].k = t[num << 1].k + t[num << 1 | 1].k;
    }
  }
  void modify(unsigned ql, unsigned qr, long long k, unsigned l = 1,
              unsigned r = sam.tot, unsigned num = 1) {
    if (ql <= l && r <= qr) {
      t[num].sum += k * t[num].k, t[num].tag += k;
      return;
    }
    pushdown(l, r, num);
    unsigned mid = (l + r) >> 1;
    if (ql <= mid) modify(ql, qr, k, l, mid, num << 1);
    if (qr > mid) modify(ql, qr, k, mid + 1, r, num << 1 | 1);
    t[num].sum = t[num << 1].sum + t[num << 1 | 1].sum;
  }
  long long Qsum(unsigned ql, unsigned qr, unsigned l = 1, unsigned r = sam.tot,
                 unsigned num = 1) {
    if (ql <= l && r <= qr) return t[num].sum;
    pushdown(l, r, num);
    unsigned mid = (l + r) >> 1;
    long long ans = 0;
    if (ql <= mid) ans += Qsum(ql, qr, l, mid, num << 1);
    if (qr > mid) ans += Qsum(ql, qr, mid + 1, r, num << 1 | 1);
    return ans;
  }
} sgt;
struct edge {
  int v, nxt;
} e[400011];
int cnt = 0, last[400011];
void adde(int u, int v) { e[++cnt].v = v, e[cnt].nxt = last[u], last[u] = cnt; }
int size[400011], fa[400011], mson[400011], dep[400011];
void dfs1(int u, int now_dep = 1) {
  size[u] = 1, dep[u] = now_dep;
  for (int i = last[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    fa[v] = u, dfs1(v, now_dep + 1);
    if (size[v] > size[mson[u]]) mson[u] = v;
    size[u] += size[v];
  }
}
int top[400011], dfn[400011], w[400011], cur = 0;
void dfs2(int u, int now_top) {
  top[u] = now_top, dfn[u] = ++cur, w[cur] = sam.len[u] - sam.len[fa[u]];
  if (mson[u]) dfs2(mson[u], now_top);
  for (int i = last[u]; i; i = e[i].nxt)
    if (e[i].v != mson[u]) dfs2(e[i].v, e[i].v);
}
void add_to_root(int u, int k) {
  while (u) sgt.modify(dfn[top[u]], dfn[u], k), u = fa[top[u]];
}
long long Qsum_to_root(int u) {
  long long res = 0;
  while (u) res += sgt.Qsum(dfn[top[u]], dfn[u]), u = fa[top[u]];
  return res;
}
int ed[400011], a[400011], b[400011];
char s[400011];
int main() {
  int n = read(), m = read();
  scanf("%s", s + 1);
  for (int i = n; i; --i) sam.insert(s[i] - 'a'), ed[i] = sam.last;
  for (int i = 2; i <= sam.tot; ++i) adde(sam.pre[i], i);
  dfs1(1), dfs2(1, 1), sgt.build(w);
  while (m--) {
    int k = read(), l = read();
    long long ans = 0;
    for (int i = 1; i <= k; ++i) a[i] = read();
    for (int i = 1; i <= l; ++i) b[i] = read(), add_to_root(ed[b[i]], 1);
    for (int i = 1; i <= k; ++i) ans += Qsum_to_root(ed[a[i]]);
    for (int i = 1; i <= l; ++i) add_to_root(ed[b[i]], -1);
    printf("%lld\n", ans);
  }
  return 0;
}