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;
inline int read() {
  char ch = getchar();
  int res = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * w;
}
const int N = 3e5 + 5, mod1 = 1e9 + 7, mod2 = 1004535809, base1 = 233,
          base2 = 403;
int n, m, sum, a[N], pow1[N], pow2[N], inv1[N], inv2[N], hsh1[N], hsh2[N],
    hsh1_[N], hsh2_[N], d[N], dfn[N], son[N], len[N], md[N], lg[N << 1],
    f[20][N], st[20][N << 1];
vector<int> e[N], down[N], up[N];
char s[N];
struct hsh {
  long long val1, val2, len;
  hsh operator+(const hsh& x) const {
    return (hsh){((long long)val1 * pow1[x.len] + x.val1) % mod1,
                 ((long long)val2 * pow2[x.len] + x.val2) % mod2, len + x.len};
  }
};
inline int Min(int x, int y) { return x < y ? x : y; }
inline int qmod(long long x, int mod) {
  if (x < 0) return x + mod;
  if (x >= mod * 2) return x % mod;
  if (x >= mod) return x - mod;
  return x;
}
inline int qpow(int x, int y, int mod) {
  int res = 1;
  while (y) {
    if (y & 1) res = qmod((long long)res * x, mod);
    x = qmod((long long)x * x, mod);
    y >>= 1;
  }
  return res;
}
void dfs1(int x) {
  d[x] = d[f[0][x]] + 1;
  hsh1[x] = qmod(qmod((long long)base1 * hsh1[f[0][x]], mod1) + s[x], mod1);
  hsh2[x] = qmod(qmod((long long)base2 * hsh2[f[0][x]], mod2) + s[x], mod2);
  hsh1_[x] =
      qmod(qmod((long long)pow1[d[x]] * s[x], mod1) + hsh1_[f[0][x]], mod1);
  hsh2_[x] =
      qmod(qmod((long long)pow2[d[x]] * s[x], mod2) + hsh2_[f[0][x]], mod2);
  dfn[x] = ++sum;
  st[0][sum] = x;
  len[x] = d[x];
  for (register int i = 1; (1 << i) < d[x]; i++)
    f[i][x] = f[i - 1][f[i - 1][x]];
  for (register int i = 0; i < e[x].size(); i++)
    if (e[x][i] != f[0][x]) {
      f[0][e[x][i]] = x;
      dfs1(e[x][i]);
      st[0][++sum] = x;
      if (len[e[x][i]] > len[son[x]]) {
        son[x] = e[x][i];
        len[x] = len[e[x][i]];
      }
    }
}
inline int lca(int x, int y) {
  int u = dfn[x], v = dfn[y];
  if (u > v) {
    int t = u;
    u = v;
    v = t;
  }
  return d[st[lg[v - u + 1]][u]] <
                 d[st[lg[v - u + 1]][v - (1 << lg[v - u + 1]) + 1]]
             ? st[lg[v - u + 1]][u]
             : st[lg[v - u + 1]][v - (1 << lg[v - u + 1]) + 1];
}
void dfs2(int x, int top) {
  md[x] = top;
  if (son[x]) dfs2(son[x], top);
  for (register int i = 0; i < e[x].size(); i++)
    if (e[x][i] != f[0][x] && e[x][i] != son[x]) dfs2(e[x][i], e[x][i]);
}
inline int kfa(int x, int k) {
  if (!k) return x;
  if (k >= d[x] || k < 0) return 0;
  x = f[lg[k]][x];
  if (!x) return 0;
  k -= (1 << lg[k]);
  return d[x] - d[md[x]] >= k ? down[md[x]][d[x] - d[md[x]] - k]
                              : up[md[x]][k - d[x] + d[md[x]]];
}
inline hsh get(int u, int v) {
  return (hsh){
      qmod(hsh1[v] - qmod((long long)hsh1[u] * pow1[d[v] - d[u]], mod1), mod1),
      qmod(hsh2[v] - qmod((long long)hsh2[u] * pow2[d[v] - d[u]], mod2), mod2),
      d[v] - d[u]};
}
inline hsh get_(int u, int v) {
  return (hsh){
      qmod((long long)qmod(hsh1_[v] - hsh1_[u], mod1) * inv1[d[u] + 1], mod1),
      qmod((long long)qmod(hsh2_[v] - hsh2_[u], mod2) * inv2[d[u] + 1], mod2),
      d[v] - d[u]};
}
signed main() {
  n = read();
  scanf("%s", s + 1);
  pow1[0] = pow2[0] = 1;
  for (register int i = 1; i <= n; i++) {
    pow1[i] = qmod((long long)pow1[i - 1] * base1, mod1);
    pow2[i] = qmod((long long)pow2[i - 1] * base2, mod2);
    inv1[i] = qpow(pow1[i], mod1 - 2, mod1);
    inv2[i] = qpow(pow2[i], mod2 - 2, mod2);
  }
  for (register int i = 1; i < n; i++) {
    int u = read(), v = read();
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1);
  dfs2(1, 1);
  for (register int i = 2; i <= sum; i++) lg[i] = lg[i >> 1] + 1;
  for (register int i = 1; i < 20; i++)
    for (register int j = 1; j + (1 << i) - 1 <= sum; j++)
      st[i][j] = d[st[i - 1][j]] < d[st[i - 1][j + (1 << i - 1)]]
                     ? st[i - 1][j]
                     : st[i - 1][j + (1 << i - 1)];
  for (register int i = 1; i <= n; i++)
    if (md[i] == i)
      for (register int j = 0, fa = i, af = i; j <= len[i] - d[i];
           j++, fa = f[0][fa], af = son[af]) {
        up[i].push_back(fa);
        down[i].push_back(af);
      }
  m = read();
  while (m--) {
    if (m == 999964) {
      int jb = 0;
    }
    int qwq = read(), QWQ = read(), qaq = read(), QAQ = read(),
        l1 = lca(qwq, QWQ), l2 = lca(qaq, QAQ),
        d1 = d[qwq] + d[QWQ] - 2 * d[l1], d2 = d[qaq] + d[QAQ] - 2 * d[l2],
        l = 0, r = Min(d1, d2), ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      hsh x = (d[qwq] - mid >= d[l1])
                  ? get_(kfa(qwq, mid + 1), qwq)
                  : (get_(f[0][l1], qwq) + get(l1, kfa(QWQ, d1 - mid))),
          y = (d[qaq] - mid >= d[l2])
                  ? get_(kfa(qaq, mid + 1), qaq)
                  : (get_(f[0][l2], qaq) + get(l2, kfa(QAQ, d2 - mid)));
      if (x.len == y.len && x.val1 == y.val1 && x.val2 == y.val2) {
        l = ans = mid + 1;
      } else
        r = mid - 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}