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;
const long long inf = 0x3f3f3f3f3f3f3f3f;
struct hld_t {
  vector<vector<int>> g;
  vector<int> sz, par;
  vector<int> h, in, rin, out, nxt;
  int t, n;
  hld_t(int _n) : n(_n) {
    h.resize(n);
    in.resize(n);
    rin.resize(n);
    out.resize(n);
    par.resize(n);
    sz.resize(n);
    g.resize(n);
    nxt.resize(n);
    t = 0;
  }
  void build(int v) {
    dfs_sz(v, v);
    hld(v, v);
  }
  void add(int first, int second) {
    g[first].push_back(second);
    g[second].push_back(first);
  }
  void dfs_sz(int v, int p) {
    sz[v] = 1, par[v] = p;
    for (int i = 0; i < g[v].size(); i++) {
      int &u = g[v][i];
      if (u == p) {
        swap(u, g[v].back());
        g[v].pop_back();
        i--;
      } else {
        h[u] = h[v] + 1;
        dfs_sz(u, v);
        sz[v] += sz[u];
        if (sz[u] > sz[g[v][0]]) {
          swap(u, g[v][0]);
        }
      }
    }
  }
  void hld(int v, int p) {
    in[v] = t++;
    rin[in[v]] = v;
    for (auto &u : g[v]) {
      nxt[u] = u == g[v][0] ? nxt[v] : u;
      hld(u, v);
    }
    out[v] = t;
  }
  int lca(int u, int v) {
    while (!inSubtree(nxt[u], v)) u = par[nxt[u]];
    while (!inSubtree(nxt[v], u)) v = par[nxt[v]];
    return in[u] < in[v] ? u : v;
  }
  bool inSubtree(int u, int v) { return in[u] <= in[v] && in[v] < out[u]; }
  vector<pair<int, int>> getPathtoAncestor(int a, int p) {
    vector<pair<int, int>> ans;
    while (nxt[a] != nxt[p]) {
      ans.emplace_back(in[nxt[a]], in[a]);
      a = par[nxt[a]];
    }
    ans.emplace_back(in[p], in[a]);
    return ans;
  }
  int dist(int a, int b) { return h[a] + h[b] - 2 * h[lca(a, b)]; }
};
template <typename T>
struct stringh {
  string s;
  vector<T> h;
  vector<T> h_base;
  T mod;
  stringh(){};
  stringh(string &s, T mod = 1) : s(s), mod(mod) {
    h.resize(s.size() + 1);
    h_base.resize(s.size() + 1);
    hash_init();
  }
  void hash_init(T prime = 0xdefaced) {
    h_base[0] = 1;
    h[0] = 0;
    for (int i = 1; i <= s.size(); i++) {
      h[i] = (h[i - 1] * prime + s[i - 1]) % mod;
      h_base[i] = (h_base[i - 1] * prime) % mod;
    }
  }
  T get(int l, int r) {
    return (h[r + 1] - (h[l] * h_base[r - l + 1]) % mod + mod) % mod;
  }
};
const int mod1 = 1073676287;
const int mod2 = 1e9 + 9;
const int mx = 312345;
struct path_h {
  stringh<long long> h1, h2;
  path_h() {}
  path_h(string &s) { h1 = stringh<long long>(s, mod1); }
} ph[2 * mx];
string s[mx];
pair<long long, long long> getKeys(vector<pair<int, int>> &paths1,
                                   vector<pair<int, int>> &paths2, int first,
                                   hld_t &t) {
  long long key1 = 0, key2 = 0;
  for (auto &a : paths1) {
    if (a.first > a.second) continue;
    auto &cur = ph[2 * t.nxt[t.rin[a.second]] ^ 1];
    int base = t.in[t.nxt[t.rin[a.second]]];
    int sz = a.second - a.first + 1;
    int tot = cur.h1.s.size();
    if (!first) break;
    if (sz > first) {
      key1 = (key1 * cur.h1.h_base[first] +
              cur.h1.get(tot - 1 - (a.second - base),
                         tot - 1 - (a.second - first + 1 - base))) %
             mod1;
      first = 0;
      break;
    }
    long long g =
        cur.h1.get(tot - 1 - (a.second - base), tot - 1 - (a.first - base));
    key1 = (key1 * cur.h1.h_base[sz] + g) % mod1;
    first -= sz;
  }
  for (auto &a : paths2) {
    if (a.first > a.second) continue;
    auto &cur = ph[2 * t.nxt[t.rin[a.second]]];
    int base = t.in[t.nxt[t.rin[a.second]]];
    int sz = a.second - a.first + 1;
    int tot = cur.h1.s.size();
    if (!first) break;
    if (sz > first) {
      key1 = (key1 * cur.h1.h_base[first] +
              cur.h1.get(a.first - base, (a.first - base) + first - 1)) %
             mod1;
      first = 0;
      break;
    }
    long long g = cur.h1.get(a.first - base, a.second - base);
    key1 = (key1 * cur.h1.h_base[sz] + g) % mod1;
    first -= sz;
  }
  return make_pair(key1, key2);
}
int main() {
  int n;
  scanf("%d", &n);
  string v;
  cin >> v;
  hld_t t(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    t.add(a, b);
  }
  t.build(0);
  for (int i = 0; i < n; i++) {
    s[t.nxt[t.rin[i]]].push_back(v[t.rin[i]]);
  }
  for (int i = 0; i < n; i++) {
    ph[2 * i] = path_h(s[i]);
    reverse(s[i].begin(), s[i].end());
    ph[2 * i + 1] = path_h(s[i]);
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b, c, d;
    scanf("%d%d", &a, &b), scanf("%d%d", &c, &d);
    a--, b--, c--, d--;
    int lo = 0, hi = min(t.dist(a, b), t.dist(c, d)) + 1;
    int lab = t.lca(a, b);
    auto paths1 = t.getPathtoAncestor(a, lab);
    auto paths2 = t.getPathtoAncestor(b, lab);
    paths2.back().first++;
    reverse(paths2.begin(), paths2.end());
    int lcd = t.lca(c, d);
    auto paths3 = t.getPathtoAncestor(c, lcd);
    auto paths4 = t.getPathtoAncestor(d, lcd);
    paths4.back().first++;
    reverse(paths4.begin(), paths4.end());
    while (hi - lo > 1) {
      int m = lo + hi >> 1;
      auto key1 = getKeys(paths1, paths2, m, t);
      auto key2 = getKeys(paths3, paths4, m, t);
      if (key1 == key2)
        lo = m;
      else
        hi = m;
    }
    auto key1 = getKeys(paths1, paths2, hi, t);
    auto key2 = getKeys(paths3, paths4, hi, t);
    printf("%d\n", key1 == key2 ? hi : lo);
  }
}