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;
template <class t>
inline void read(t& res) {
  char ch;
  while (ch = getchar(), !isdigit(ch))
    ;
  res = ch ^ 48;
  while (ch = getchar(), isdigit(ch)) res = res * 10 + (ch ^ 48);
}
template <class t>
inline void print(t x) {
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
const int e = 3e5 + 5, z1 = 29, z2 = 31, h1 = 1e9 + 7, h2 = 1e9 + 9;
struct node {
  int s1, s2, len;
  node() {}
  node(int _s1, int _s2, int _len) : s1(_s1), s2(_s2), len(_len) {}
} up[e], down[e];
vector<int> a[e], b[e], g[e];
int n, top[e], son[e], m, f[20][e], dep[e], mx[e];
int p1[e], p2[e], fa[e], dfn[e], id[e << 1], tim, logn[e << 1], c[21][e << 1];
int q1[e], q2[e], inv1, inv2;
char s[e];
inline bool operator==(node a, node b) { return a.s1 == b.s1 && a.s2 == b.s2; }
inline node operator+(node a, node b) {
  return node(((long long)a.s1 * p1[b.len] + b.s1) % h1,
              ((long long)a.s2 * p2[b.len] + b.s2) % h2, a.len + b.len);
}
inline int ksm(int x, int y, int mod) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % mod;
    y >>= 1;
    x = (long long)x * x % mod;
  }
  return res;
}
inline void dfs1(int u, int pa) {
  int i;
  fa[u] = pa;
  dep[u] = dep[pa] + 1;
  mx[u] = dep[u];
  s[u] -= 'a' - 1;
  dfn[u] = ++tim;
  id[tim] = u;
  for (i = 0; i < 18; i++) f[i + 1][u] = f[i][f[i][u]];
  node a = down[pa], b = up[pa];
  down[u] = node(((long long)z1 * a.s1 + s[u]) % h1,
                 ((long long)z2 * a.s2 + s[u]) % h2, a.len + 1);
  up[u] = node((b.s1 + (long long)p1[b.len] * s[u]) % h1,
               (b.s2 + (long long)p2[b.len] * s[u]) % h2, b.len + 1);
  for (auto v : g[u])
    if (v != pa) {
      f[0][v] = u;
      dfs1(v, u);
      if (mx[v] > mx[u]) mx[u] = mx[v], son[u] = v;
      id[++tim] = u;
    }
}
inline void dfs2(int u, int pa, int fi) {
  top[u] = fi;
  if (son[u]) dfs2(son[u], u, fi);
  for (auto v : g[u])
    if (v != pa && v != son[u]) dfs2(v, u, v);
}
inline int cmp(int x, int y) { return dep[x] < dep[y] ? x : y; }
inline void init() {
  int i, j;
  logn[0] = -1;
  for (i = 1; i <= tim; i++) logn[i] = logn[i >> 1] + 1, c[0][i] = id[i];
  for (j = 1; (1 << j) <= tim; j++)
    for (i = 1; i + (1 << j) - 1 <= tim; i++)
      c[j][i] = cmp(c[j - 1][i], c[j - 1][i + (1 << j - 1)]);
}
inline int lca(int l, int r) {
  l = dfn[l];
  r = dfn[r];
  if (l > r) swap(l, r);
  int k = logn[r - l + 1];
  return cmp(c[k][l], c[k][r - (1 << k) + 1]);
}
inline int anc(int x, int k) {
  if (!k) return x;
  int y = logn[k], to;
  x = f[y][x];
  k -= (1 << y);
  to = top[x];
  if (dep[to] <= dep[x] - k)
    return b[to][dep[x] - k - dep[to]];
  else
    return a[to][dep[to] - dep[x] + k];
}
inline int sub1(int x, int y) {
  (x -= y) < 0 && (x += h1);
  return x;
}
inline int sub2(int x, int y) {
  (x -= y) < 0 && (x += h2);
  return x;
}
inline node query_up(int x, int y) {
  node a = up[x], b = up[y];
  return node((long long)sub1(a.s1, b.s1) * q1[b.len] % h1,
              (long long)sub2(a.s2, b.s2) * q2[b.len] % h2, a.len - b.len);
}
inline node query_down(int x, int y) {
  node a = down[y], b = down[x];
  return node(sub1(a.s1, (long long)b.s1 * p1[a.len - b.len] % h1),
              sub2(a.s2, (long long)b.s2 * p2[a.len - b.len] % h2),
              a.len - b.len);
}
inline node calc(int x, int y, int z, int len) {
  int d1 = dep[x] - dep[z] + 1, d2 = dep[x] + dep[y] - 2 * dep[z] + 1;
  if (len <= d1) {
    int u = anc(x, len - 1);
    return query_up(x, fa[u]);
  } else {
    int u = anc(y, d2 - len);
    return query_up(x, fa[z]) + query_down(z, u);
  }
}
inline int query(int x, int y, int u, int v) {
  int z = lca(x, y), w = lca(u, v), res = 0, l = 1,
      r = min(dep[x] + dep[y] - 2 * dep[z], dep[u] + dep[v] - 2 * dep[w]) + 1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (calc(x, y, z, mid) == calc(u, v, w, mid))
      res = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return res;
}
int main() {
  read(n);
  int i, x, y, u, v, j;
  scanf("%s", s + 1);
  for (i = 1; i < n; i++) {
    read(x);
    read(y);
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  inv1 = ksm(z1, h1 - 2, h1);
  inv2 = ksm(z2, h2 - 2, h2);
  p1[0] = p2[0] = q1[0] = q2[0] = 1;
  for (i = 1; i <= n; i++) {
    p1[i] = (long long)p1[i - 1] * z1 % h1;
    p2[i] = (long long)p2[i - 1] * z2 % h2;
    q1[i] = (long long)q1[i - 1] * inv1 % h1;
    q2[i] = (long long)q2[i - 1] * inv2 % h2;
  }
  dfs1(1, 0);
  dfs2(1, 0, 1);
  init();
  for (i = 1; i <= n; i++)
    if (top[i] == i) {
      int x = i;
      for (j = 0; j <= mx[i] - dep[i]; j++) a[i].emplace_back(x), x = fa[x];
      x = i;
      while (x) b[i].emplace_back(x), x = son[x];
    }
  read(m);
  while (m--) {
    read(x);
    read(y);
    read(u);
    read(v);
    print(query(x, y, u, v));
    putchar('\n');
  }
  return 0;
}