Problem: Misha has a tree with characters written on the vertices. He can choose two vertices s and t of this tree and write down characters of vertices lying on a path from s to t. We'll say that such string corresponds to pair (s, t).

Misha has m queries of type: you are given 4 vertices a, b, c, d; you need to find the largest common prefix of the strings that correspond to pairs (a, b) and (c, d). Your task is to help him.

Input

The first line contains integer n (1 ≤ n ≤ 300 000) — the number of vertices in the tree.

Next follows a line consisting of n small English letters. The i-th character of the string corresponds to the character written on the i-th vertex. 

Next n - 1 lines contain information about edges. An edge is defined by a pair of integers u, v (1 ≤ u, v ≤ n, u ≠ v), separated by spaces.

The next line contains integer m (1 ≤ m ≤ 1 000 000) — the number of queries.

Next m lines contain information about queries. A query is defined by four integers a, b, c, d (1 ≤ a, b, c, d ≤ n), separated by spaces.

Output

For each query print the length of the largest common prefix on a separate line.

Examples

Input

6
bbbabb
2 1
3 2
4 3
5 2
6 5
6
2 5 3 1
1 5 2 3
5 6 5 6
6 3 4 1
6 2 3 4
2 2 4 5


Output

2
2
2
0
1
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int P1 = 998244353, P2 = 1000000007, N = 300005;
struct Hash {
  ll h1, h2;
} B = {19260817, 101}, h1[N], h2[N];
inline Hash operator+(Hash a, Hash b) {
  return {(a.h1 + b.h1) % P1, (a.h2 + b.h2) % P2};
}
inline Hash operator-(Hash a, Hash b) {
  return {(a.h1 - b.h1 + P1) % P1, (a.h2 - b.h2 + P2) % P2};
}
inline Hash operator*(Hash a, Hash b) {
  return {a.h1 * b.h1 % P1, a.h2 * b.h2 % P2};
}
inline bool operator==(Hash a, Hash b) { return a.h1 == b.h1 && a.h2 == b.h2; }
int n, m;
char s[N];
Hash pw[N];
vector<int> v[N];
template <typename T>
void read(T &num) {
  num = 0;
  int ch = getchar();
  T fl = 1;
  while (!isdigit(ch) && ch != 45) ch = getchar();
  if (ch == 45) fl = -1, ch = getchar();
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - 48);
    ch = getchar();
  }
  num *= fl;
}
template <typename T, typename... Arg>
void read(T &num, Arg &...rest) {
  read(num);
  read(rest...);
}
int sz[N], fa[N], dep[N], hson[N], dfn[N], idx, mp[N], top[N];
void dfs(int pos) {
  sz[pos] = 1;
  dep[pos] = dep[fa[pos]] + 1;
  for (auto &i : v[pos])
    if (i != fa[pos]) {
      fa[i] = pos;
      dfs(i);
      sz[pos] += sz[i];
      if (sz[i] > sz[hson[pos]]) hson[pos] = i;
    }
}
void dfs(int pos, int tp) {
  top[pos] = tp;
  dfn[pos] = ++idx;
  mp[idx] = pos;
  if (hson[pos]) dfs(hson[pos], tp);
  for (auto &i : v[pos])
    if (i != fa[pos] && i != hson[pos]) dfs(i, i);
}
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;
}
vector<pair<int, int>> get(int x, int y) {
  int z = lca(x, y);
  vector<pair<int, int>> ret, tmp;
  while (dep[top[x]] > dep[z]) ret.push_back({x, top[x]}), x = fa[top[x]];
  ret.push_back({x, z});
  while (dep[top[y]] > dep[z]) tmp.push_back({top[y], y}), y = fa[top[y]];
  if (y != z) tmp.push_back({hson[z], y});
  while (!tmp.empty()) ret.push_back(tmp.back()), tmp.pop_back();
  return ret;
}
inline Hash geth(bool up, int pos, int len) {
  if (up) return h2[pos - len + 1] - h2[pos + 1] * pw[len];
  return h1[pos + len - 1] - h1[pos - 1] * pw[len];
}
int main() {
  ios::sync_with_stdio(false);
  read(n);
  scanf("%s", s + 1);
  pw[0] = {1, 1};
  pw[1] = B;
  for (int i = 2; i <= n; i++) pw[i] = pw[i - 1] * B;
  int t1, t2;
  for (int i = 1; i < n; i++) {
    read(t1, t2);
    v[t1].push_back(t2);
    v[t2].push_back(t1);
  }
  dfs(1);
  dfs(1, 1);
  for (int i = 1; i <= n; i++) h1[i] = h1[i - 1] * B + Hash{s[mp[i]], s[mp[i]]};
  for (int i = n; i > 0; --i) h2[i] = h2[i + 1] * B + Hash{s[mp[i]], s[mp[i]]};
  read(m);
  while (m--) {
    int a, b, c, d, ans = 0, p1 = 0, p2 = 0;
    read(a, b, c, d);
    auto aa = get(a, b), bb = get(c, d);
    while (p1 < int(aa.size()) && p2 < int(bb.size())) {
      int ida1 = dfn[aa[p1].first], ida2 = dfn[aa[p1].second];
      int idb1 = dfn[bb[p2].first], idb2 = dfn[bb[p2].second];
      bool oa = ida1 > ida2, ob = idb1 > idb2;
      int la = abs(ida1 - ida2) + 1, lb = abs(idb1 - idb2) + 1,
          len = min(la, lb);
      Hash ha = geth(oa, ida1, len), hb = geth(ob, idb1, len);
      if (ha == hb) {
        ans += len;
        if (len == la)
          ++p1;
        else
          aa[p1].first = mp[ida1 + (oa ? -1 : 1) * len];
        if (len == lb)
          ++p2;
        else
          bb[p2].first = mp[idb1 + (ob ? -1 : 1) * len];
      } else {
        int l = 1, r = len, tmp = 0;
        while (l <= r) {
          int mid = (l + r) >> 1;
          Hash ha = geth(oa, ida1, mid), hb = geth(ob, idb1, mid);
          if (ha == hb)
            tmp = mid, l = mid + 1;
          else
            r = mid - 1;
        }
        ans += tmp;
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}