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() {
  int res = 0;
  char c;
  bool zf = 0;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    zf = 1;
  else
    res = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + c - '0';
  if (zf) return -res;
  return res;
}
int dta[300005];
vector<int> G[300005];
int power1[300005], power2[300005];
int inv1[300005], inv2[300005];
int Hash1[300005][2], Hash2[300005][2];
vector<int> _up[300005], _down[300005];
int dep[300005], maxdep[300005], fa[300005], size[300005], son1[300005],
    son2[300005], top1[300005], top2[300005];
int ffa[300005][25];
int highbit[300005];
inline int ksm(int x, int y, int mod) {
  int res = 1;
  while (y) {
    if (y & 1) {
      res = 1ll * res * x % mod;
    }
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
void dfs1(int x) {
  size[x] = 1;
  for (register int i = 0; i < G[x].size(); ++i) {
    int v = G[x][i];
    if (dep[v]) {
      continue;
    }
    dep[v] = dep[x] + 1;
    maxdep[v] = 1;
    fa[v] = x;
    dfs1(v);
    size[x] += size[v];
    maxdep[x] = max(maxdep[x], maxdep[v] + 1);
    if (size[v] > size[son1[x]]) {
      son1[x] = v;
    }
    if (maxdep[v] > maxdep[son2[x]]) {
      son2[x] = v;
    }
  }
  return;
}
void dfs2(int x) {
  if (!son1[x]) {
    return;
  }
  if (son1[x] == son2[x]) {
    top1[son1[x]] = top1[x];
    top2[son2[x]] = top2[x];
    dfs2(son1[x]);
  } else {
    top1[son1[x]] = top1[x];
    dfs2(top2[son1[x]] = son1[x]);
    top2[son2[x]] = top2[x];
    dfs2(top1[son2[x]] = son2[x]);
  }
  for (register int i = 0; i < G[x].size(); ++i) {
    int v = G[x][i];
    if (v == fa[x] || v == son1[x] || v == son2[x]) {
      continue;
    }
    dfs2(top1[v] = top2[v] = v);
  }
  return;
}
inline int LCA(int x, int y) {
  while (top1[x] != top1[y]) {
    if (dep[top1[x]] < dep[top1[y]]) {
      swap(x, y);
    }
    x = fa[top1[x]];
  }
  if (dep[x] > dep[y]) {
    swap(x, y);
  }
  return x;
}
void getfa(int x) {
  ffa[x][0] = fa[x];
  for (register int i = 1; i <= 19; ++i) {
    ffa[x][i] = ffa[ffa[x][i - 1]][i - 1];
  }
  for (register int i = 0; i < G[x].size(); ++i) {
    int v = G[x][i];
    if (v == fa[x]) {
      continue;
    }
    getfa(v);
  }
  return;
}
inline void up_and_down(int n) {
  for (register int i = 1; i <= n; ++i) {
    if (top2[i] == i) {
      int now = i;
      while (now) {
        _down[i].push_back(now);
        now = son2[now];
      }
      now = i;
      while (now && _up[i].size() + 1 <= _down[i].size()) {
        _up[i].push_back(now);
        now = fa[now];
      }
    }
  }
  return;
}
inline int kthfa(int x, int k) {
  if (dep[x] <= k) {
    return 0;
  }
  if (!k) {
    return x;
  }
  int tmp = highbit[k];
  x = ffa[x][tmp];
  k -= (1 << tmp);
  if (!k) {
    return x;
  }
  int cz = dep[x] - dep[top2[x]];
  if (cz >= k) {
    return _down[top2[x]][cz - k];
  }
  return _up[top2[x]][k - cz];
}
inline int dist(int x, int y) {
  int l = LCA(x, y);
  return dep[x] + dep[y] - 2 * dep[l] + 1;
}
void get_Hash(int now) {
  Hash1[now][0] =
      (1ll * Hash1[fa[now]][0] * 19260817 % 1000000007 + dta[now]) % 1000000007;
  Hash1[now][1] =
      (Hash1[fa[now]][1] + 1ll * dta[now] * power1[dep[now] - 1] % 1000000007) %
      1000000007;
  Hash2[now][0] =
      (1ll * Hash2[fa[now]][0] * 19491001 % 1000000009 + dta[now]) % 1000000009;
  Hash2[now][1] =
      (Hash2[fa[now]][1] + 1ll * dta[now] * power2[dep[now] - 1] % 1000000009) %
      1000000009;
  for (register int i = 0; i < G[now].size(); ++i) {
    int v = G[now][i];
    if (v == fa[now]) {
      continue;
    }
    get_Hash(v);
  }
  return;
}
inline int _get(int x, int y, int opt) {
  if (dep[x] > dep[y]) {
    if (opt == 1) {
      return 1ll * ((Hash1[x][1] - Hash1[fa[y]][1] + 1000000007) % 1000000007) *
             inv1[dep[y] - 1] % 1000000007;
    }
    return 1ll * ((Hash2[x][1] - Hash2[fa[y]][1] + 1000000009) % 1000000009) *
           inv2[dep[y] - 1] % 1000000009;
  }
  if (opt == 1) {
    return (Hash1[y][0] -
            (1ll * Hash1[fa[x]][0] * power1[dep[y] - dep[x] + 1] % 1000000007) +
            1000000007) %
           1000000007;
  }
  return (Hash2[y][0] -
          (1ll * Hash2[fa[x]][0] * power2[dep[y] - dep[x] + 1] % 1000000009) +
          1000000009) %
         1000000009;
}
signed main() {
  int n = read();
  for (register int i = 0; (1 << i) <= n; ++i) {
    highbit[1 << i] = i;
  }
  for (register int i = 2; i <= n; ++i)
    if (!highbit[i]) {
      highbit[i] = highbit[i - 1];
    }
  string s;
  cin >> s;
  for (register int i = 1; i <= n; ++i) {
    dta[i] = s[i - 1] - 'a';
  }
  for (register int i = 1; i < n; ++i) {
    int x = read(), y = read();
    G[x].push_back(y), G[y].push_back(x);
  }
  power1[0] = power2[0] = 1;
  for (register int i = 1; i <= n; ++i) {
    power1[i] = 1ll * power1[i - 1] * 19260817 % 1000000007;
    power2[i] = 1ll * power2[i - 1] * 19491001 % 1000000009;
  }
  int _inv1 = ksm(19260817, 1000000007 - 2, 1000000007),
      _inv2 = ksm(19491001, 1000000009 - 2, 1000000009);
  inv1[0] = inv2[0] = 1;
  for (register int i = 1; i <= n; ++i) {
    inv1[i] = 1ll * inv1[i - 1] * _inv1 % 1000000007;
    inv2[i] = 1ll * inv2[i - 1] * _inv2 % 1000000009;
  }
  dfs1(dep[1] = maxdep[1] = 1);
  dfs2(top1[1] = top2[1] = 1);
  getfa(1);
  up_and_down(n);
  get_Hash(1);
  int m = read();
  while (m--) {
    int a = read(), b = read(), c = read(), d = read();
    int d1 = dist(a, b), d2 = dist(c, d), lca1 = LCA(a, b), lca2 = LCA(c, d),
        l = 1, r = min(d1, d2), ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      int _hash1 =
              ((dep[a] - dep[lca1] + 1 >= mid)
                   ? _get(a, kthfa(a, mid - 1), 1)
                   : (((1ll * _get(a, lca1, 1) *
                        power1[mid - (dep[a] - dep[lca1] + 1)] % 1000000007) +
                       (_get(lca1, kthfa(b, d1 - mid), 1) -
                        1ll * dta[lca1] *
                            power1[mid - (dep[a] - dep[lca1] + 1)] %
                            1000000007 +
                        1000000007) %
                           1000000007) %
                      1000000007)),
          _hash2 =
              ((dep[c] - dep[lca2] + 1 >= mid)
                   ? _get(c, kthfa(c, mid - 1), 1)
                   : (((1ll * _get(c, lca2, 1) *
                        power1[mid - (dep[c] - dep[lca2] + 1)] % 1000000007) +
                       (_get(lca2, kthfa(d, d2 - mid), 1) -
                        1ll * dta[lca2] *
                            power1[mid - (dep[c] - dep[lca2] + 1)] %
                            1000000007 +
                        1000000007) %
                           1000000007) %
                      1000000007));
      if (_hash1 != _hash2) {
        r = mid - 1;
        continue;
      }
      _hash1 =
          ((dep[a] - dep[lca1] + 1 >= mid)
               ? _get(a, kthfa(a, mid - 1), 2)
               : (((1ll * _get(a, lca1, 2) *
                    power2[mid - (dep[a] - dep[lca1] + 1)] % 1000000009) +
                   (_get(lca1, kthfa(b, d1 - mid), 2) -
                    1ll * dta[lca1] * power2[mid - (dep[a] - dep[lca1] + 1)] %
                        1000000009 +
                    1000000009) %
                       1000000009) %
                  1000000009)),
      _hash2 =
          ((dep[c] - dep[lca2] + 1 >= mid)
               ? _get(c, kthfa(c, mid - 1), 2)
               : (((1ll * _get(c, lca2, 2) *
                    power2[mid - (dep[c] - dep[lca2] + 1)] % 1000000009) +
                   (_get(lca2, kthfa(d, d2 - mid), 2) -
                    1ll * dta[lca2] * power2[mid - (dep[c] - dep[lca2] + 1)] %
                        1000000009 +
                    1000000009) %
                       1000000009) %
                  1000000009));
      if (_hash1 != _hash2) {
        r = mid - 1;
        continue;
      }
      ans = mid;
      l = mid + 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}