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 <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline char Get() {
  static char c;
  c = getchar();
  while (!isalpha(c)) c = getchar();
  return c;
}
inline void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 300005;
int To[N << 1], Ne[N << 1], He[N], _;
inline void add(int x, int y) {
  ++_;
  To[_] = y, Ne[_] = He[x], He[x] = _;
  ++_;
  To[_] = x, Ne[_] = He[y], He[y] = _;
}
char a[N], s[N << 1];
int n, fa[N], dpt[N], size[N], son[N], top[N], id[N], tl[N], tr[N], pos[N],
    Time;
inline void dfs1(int x) {
  dpt[x] = dpt[fa[x]] + 1, size[x] = 1;
  for (int y, p = He[x]; p; p = Ne[p])
    if (fa[x] ^ (y = To[p])) {
      fa[y] = x, dfs1(y), size[x] += size[y];
      if (size[y] > size[son[x]]) son[x] = y;
    }
}
inline void dfs2(int x) {
  id[x] = tl[x] = ++Time;
  pos[Time] = x;
  s[Time] = a[x], s[(n << 1 | 1) - Time] = a[x];
  if (son[x]) {
    top[son[x]] = top[x], dfs2(son[x]);
    for (int y, p = He[x]; p; p = Ne[p])
      if (!top[y = To[p]]) top[y] = y, dfs2(y);
  }
  tr[x] = Time;
}
inline int LCA(int x, int y) {
  while (top[x] ^ top[y])
    if (dpt[top[x]] > dpt[top[y]])
      x = fa[top[x]];
    else
      y = fa[top[y]];
  return dpt[x] <= dpt[y] ? x : y;
}
int sa[N << 1], rk[N << 1], h[N << 1], st[20][N << 1], Log[N << 1];
inline int LCP(int x, int y) {
  if (x == y) return (n << 1 | 1) - x;
  x = rk[x], y = rk[y];
  if (x > y) swap(x, y);
  ++x;
  static int t;
  t = Log[y - x + 1];
  return min(st[t][x], st[t][y - (1 << t) + 1]);
}
struct SuffixArray {
  int x[N << 1], y[N << 1], c[N << 1], l, a[N << 1];
  inline void getsa(int n) {
    register int i;
    int j, m;
    l = max('z' - 'a' + 1, n);
    for (i = 1; i <= n; ++i) rk[i] = s[i] - 'a' + 1;
    for (j = 1;; j <<= 1) {
      for (i = 1; i <= n; ++i)
        x[i] = rk[i], y[i] = (i + j <= n) ? rk[i + j] : 0;
      memset(c, 0, (l + 1) << 2);
      for (i = 1; i <= n; ++i) ++c[y[i]];
      for (i = 1; i <= l; ++i) c[i] += c[i - 1];
      for (i = 1; i <= n; ++i) a[c[y[i]]--] = i;
      memset(c, 0, (l + 1) << 2);
      for (i = 1; i <= n; ++i) ++c[x[i]];
      for (i = 1; i <= l; ++i) c[i] += c[i - 1];
      for (i = n; i; --i) sa[c[x[a[i]]]--] = a[i];
      m = 0;
      for (i = 1; i <= n; ++i) {
        rk[sa[i]] =
            (x[sa[i]] != x[sa[i - 1]] || y[sa[i]] != y[sa[i - 1]]) ? ++m : m;
      }
      if (m == n) break;
    }
    int now = 0;
    for (i = 1; i <= n; ++i) {
      if (now) --now;
      while (s[i + now] == s[sa[rk[i] - 1] + now]) ++now;
      h[rk[i]] = st[0][rk[i]] = now;
    }
    for (i = 1; i <= n; ++i) {
      Log[i] = Log[i - 1];
      if ((1 << Log[i] + 1) < i) ++Log[i];
    }
    for (i = 1; i <= Log[n]; ++i)
      for (j = 1; j <= n; ++j)
        if (j + (1 << i) - 1 <= n)
          st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << i - 1)]);
  }
} SA;
int l1[N], r1[N], n1;
inline void get1(int x, int y) {
  int z = LCA(x, y);
  n1 = 0;
  while (top[x] ^ top[z])
    ++n1, l1[n1] = (n << 1 | 1) - id[x], r1[n1] = (n << 1 | 1) - id[top[x]],
          x = fa[top[x]];
  if (id[z] < id[x])
    ++n1, l1[n1] = (n << 1 | 1) - id[x], r1[n1] = (n << 1 | 1) - (id[z] + 1);
  int l = n1;
  while (top[y] ^ top[z])
    ++n1, l1[n1] = id[top[y]], r1[n1] = id[y], y = fa[top[y]];
  ++n1, l1[n1] = id[z], r1[n1] = id[y];
  reverse(l1 + l + 1, l1 + n1 + 1), reverse(r1 + l + 1, r1 + n1 + 1);
}
int l2[N], r2[N], n2;
inline void get2(int x, int y) {
  int z = LCA(x, y);
  n2 = 0;
  while (top[x] ^ top[z])
    ++n2, l2[n2] = (n << 1 | 1) - id[x], r2[n2] = (n << 1 | 1) - id[top[x]],
          x = fa[top[x]];
  if (id[z] < id[x])
    ++n2, l2[n2] = (n << 1 | 1) - id[x], r2[n2] = (n << 1 | 1) - (id[z] + 1);
  int l = n2;
  while (top[y] ^ top[z])
    ++n2, l2[n2] = id[top[y]], r2[n2] = id[y], y = fa[top[y]];
  ++n2, l2[n2] = id[z], r2[n2] = id[y];
  reverse(l2 + l + 1, l2 + n2 + 1), reverse(r2 + l + 1, r2 + n2 + 1);
}
int ans, now1, now2;
inline int solve() {
  int len, len1, len2;
  ans = 0, now1 = now2 = 1;
  while (now1 <= n1 && now2 <= n2) {
    len1 = r1[now1] - l1[now1] + 1;
    len2 = r2[now2] - l2[now2] + 1;
    len = LCP(l1[now1], l2[now2]);
    if (len > len1) len = len1;
    if (len > len2) len = len2;
    ans += len;
    if (len < len1 && len < len2) break;
    if (len1 == len2) {
      ++now1, ++now2;
      continue;
    }
    if (len1 < len2)
      ++now1, l2[now2] += len;
    else
      ++now2, l1[now1] += len;
  }
  return ans;
}
int main() {
  int i, x, y, q;
  read(n);
  for (i = 1; i <= n; ++i) a[i] = Get();
  for (i = 1; i < n; ++i) read(x), read(y), add(x, y);
  dfs1(1), top[1] = 1, dfs2(1);
  s[0] = s[n << 1 | 1] = '\0', SA.getsa(n << 1);
  read(q);
  int qa, qb, qc, qd;
  while (q--) {
    read(qa), read(qb), read(qc), read(qd);
    get1(qa, qb);
    get2(qc, qd);
    write(solve()), putchar('\n');
  }
  return 0;
}