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 int Maxn = 300005, p = 1e9 + 9, seed = 29;
int n, q, ans, ct, cnt, inv_seed, head[Maxn], siz[Maxn], dfn[Maxn], bel[Maxn],
    bot[Maxn], son[Maxn], top[Maxn], fa[Maxn], dep[Maxn];
long long Hash1[Maxn], Hash2[Maxn], Pow[Maxn], inv[Maxn];
string str;
struct edg {
  int nxt, to;
} edge[2 * Maxn];
void add(int x, int y) {
  edge[++cnt] = (edg){head[x], y};
  head[x] = cnt;
}
long long fast_pow(long long x, long long y) {
  long long ans = 1, now = x;
  while (y) {
    if (y & 1) ans = ans * now % p;
    now = now * now % p;
    y >>= 1;
  }
  return ans;
}
void dfs1(int u, int f) {
  siz[u] = 1;
  dep[u] = dep[f] + 1;
  fa[u] = f;
  Hash1[u] = (Hash1[f] * seed + str[u - 1] - 'a') % p;
  Hash2[u] = (Hash2[f] + (str[u - 1] - 'a') * Pow[dep[u]] % p) % p;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int to = edge[i].to;
    if (to != f) {
      dfs1(to, u);
      siz[u] += siz[to];
      if (siz[to] > siz[son[u]]) son[u] = to;
    }
  }
}
void dfs2(int u, int tp) {
  bot[u] = u;
  dfn[u] = ++ct;
  bel[ct] = u;
  top[u] = tp;
  if (son[u]) dfs2(son[u], tp), bot[u] = bot[son[u]];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int to = edge[i].to;
    if (to != fa[u] && to != son[u]) dfs2(to, to), bot[to] = bot[son[to]];
  }
}
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] ? y : x;
}
int get_hash(int pos, int range, int type) {
  if (type) {
    int pos2 = bel[dfn[pos] + range - 1];
    return (Hash1[pos2] - Hash1[fa[pos]] * Pow[range] % p + p) % p;
  } else {
    int pos2 = bel[dfn[pos] - range + 1];
    return (Hash2[pos] - Hash2[fa[pos2]] + p) * inv[dep[fa[pos2]] + 1] % p;
  }
}
vector<int> Ve1, Ve2;
int work(int a1, int b1, int a2, int b2) {
  Ve1.clear(), Ve2.clear();
  ans = 0;
  int lca1 = lca(a1, b1), lca2 = lca(a2, b2);
  int tmpb1 = b1, tmpb2 = b2;
  while (top[tmpb1] != top[lca1])
    Ve1.push_back(top[tmpb1]), tmpb1 = fa[top[tmpb1]];
  while (top[tmpb2] != top[lca2])
    Ve2.push_back(top[tmpb2]), tmpb2 = fa[top[tmpb2]];
  reverse(Ve1.begin(), Ve1.end());
  reverse(Ve2.begin(), Ve2.end());
  Ve1.push_back(n + 1);
  Ve2.push_back(n + 1);
  int pnt1 = 0, pnt2 = 0;
  bool type1 = false, type2 = false;
  if (a1 == lca1) type1 = true;
  if (a2 == lca2) type2 = true;
  while (true) {
    int now = 0x3f3f3f3f;
    if (type1)
      now = min(now, min(dep[b1] - dep[a1], dep[Ve1[pnt1]] - 1 - dep[a1]));
    else
      now = min(now, min(dep[a1] - dep[lca1], dep[a1] - dep[top[a1]]));
    if (type2)
      now = min(now, min(dep[b2] - dep[a2], dep[Ve2[pnt2]] - 1 - dep[a2]));
    else
      now = min(now, min(dep[a2] - dep[lca2], dep[a2] - dep[top[a2]]));
    now++;
    if (get_hash(a1, now, type1) == get_hash(a2, now, type2)) {
      ans += now;
      if ((type1 && bel[dfn[a1] + now - 1] == b1) ||
          (!type1 && bel[dfn[a1] - now + 1] == b1) ||
          (type2 && bel[dfn[a2] + now - 1] == b2) ||
          (!type2 && bel[dfn[a2] - now + 1] == b2))
        return ans;
      if (type1) {
        if (bel[dfn[a1] + now - 1] != fa[Ve1[pnt1]])
          a1 = bel[dfn[a1] + now];
        else
          a1 = Ve1[pnt1++];
      } else {
        if (bel[dfn[a1] - now + 1] == lca1) {
          type1 = true;
          if (fa[Ve1[0]] == lca1)
            a1 = Ve1[pnt1++];
          else
            a1 = son[lca1];
        } else
          a1 = fa[bel[dfn[a1] - now + 1]];
      }
      if (type2) {
        if (bel[dfn[a2] + now - 1] != fa[Ve2[pnt2]])
          a2 = bel[dfn[a2] + now];
        else
          a2 = Ve2[pnt2++];
      } else {
        if (bel[dfn[a2] - now + 1] == lca2) {
          type2 = true;
          if (fa[Ve2[0]] == lca2)
            a2 = Ve2[pnt2++];
          else
            a2 = son[lca2];
        } else
          a2 = fa[bel[dfn[a2] - now + 1]];
      }
    } else {
      int l = 0, r = now - 1;
      while (l + 1 <= r) {
        int mid = (l + r + 1) >> 1;
        if (get_hash(a1, mid, type1) == get_hash(a2, mid, type2))
          l = mid;
        else
          r = mid - 1;
      }
      return ans + l;
    }
    if (a1 == lca1) type1 = true;
    if (a2 == lca2) type2 = true;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  dep[n + 1] = 0x3f3f3f3f;
  inv_seed = fast_pow(seed, p - 2);
  Pow[0] = 1, inv[0] = 1;
  for (int i = 1; i <= n; i++)
    Pow[i] = Pow[i - 1] * seed % p, inv[i] = inv[i - 1] * inv_seed % p;
  cin >> str;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    printf("%d\n", work(a, b, c, d));
  }
  return 0;
}