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>
bool ckmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
bool ckmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
const int N = 300005;
inline int qpow(int n, int k, int mod) {
  int res = 1;
  for (; k; k >>= 1, n = 1ll * n * n % mod)
    if (k & 1) res = 1ll * n * res % mod;
  return res;
}
int n, m, h[N];
char str[N];
struct edge {
  int nx, to;
} e[N << 1];
int hed[N], et;
inline void adde(int u, int v) {
  e[++et].nx = hed[u], e[et].to = v, hed[u] = et;
}
int dep[N], son[N], top[N], len[N], fa[N], lg[N << 1], bz[20][N];
int pool[N << 2], *up[N], *dw[N], *mem = pool;
long long d1[N], d2[N], u1[N], u2[N], pw1[N], pw2[N], ipw1[N], ipw2[N];
int ST[20][N << 1], dfn[N], tmr;
void dfs1(int u, int ft) {
  dep[u] = dep[ft] + 1, bz[0][u] = fa[u] = ft, ST[0][dfn[u] = ++tmr] = u;
  for (int i = 1; 1 << i < dep[u]; ++i) bz[i][u] = bz[i - 1][bz[i - 1][u]];
  u1[u] = (u1[ft] * 31 + h[u]) % 998244353;
  u2[u] = (u2[ft] * 37 + h[u]) % 1004535809;
  d1[u] = (d1[ft] + pw1[dep[u] - 1] * h[u]) % 998244353;
  d2[u] = (d2[ft] + pw2[dep[u] - 1] * h[u]) % 1004535809;
  for (int i = hed[u]; i; i = e[i].nx) {
    int v = e[i].to;
    if (v == ft) continue;
    dfs1(v, u), ST[0][++tmr] = u;
    if (len[v] > len[son[u]]) son[u] = v;
  }
  len[u] = len[son[u]] + 1;
}
void dfs2(int u, int tp) {
  top[u] = tp;
  if (son[u]) dfs2(son[u], tp);
  for (int i = hed[u]; i; i = e[i].nx) {
    int v = e[i].to;
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
  if (top[u] == u) {
    up[u] = mem;
    for (int i = 0, j = u; i <= len[u] && j; j = fa[j], ++i) *mem++ = j;
    dw[u] = mem;
    for (int j = u; j; j = son[j]) *mem++ = j;
  }
}
inline int kthanc(int x, int k) {
  if (!k) return x;
  int y = top[bz[lg[k]][x]];
  if (dep[x] - dep[y] >= k)
    return dw[y][dep[x] - dep[y] - k];
  else
    return up[y][k - dep[x] + dep[y]];
}
inline int LCA(int x, int y) {
  if (dfn[x] > dfn[y]) x ^= y ^= x ^= y;
  int t = lg[dfn[y] - dfn[x] + 1], t1 = ST[t][dfn[x]],
      t2 = ST[t][dfn[y] - (1 << t) + 1];
  return dep[t1] < dep[t2] ? t1 : t2;
}
inline long long calc1(long long* h, int v, int u) {
  return (h[v] - h[fa[u]] * pw1[dep[v] - dep[fa[u]]] % 998244353 + 998244353) %
         998244353;
}
inline long long calc2(long long* h, int v, int u) {
  return (h[v] - h[fa[u]] * pw2[dep[v] - dep[fa[u]]] % 1004535809 +
          1004535809) %
         1004535809;
}
inline long long calc3(long long* h, int v, int u) {
  return (h[v] - h[u] + 998244353) % 998244353 * ipw1[dep[u]] % 998244353;
}
inline long long calc4(long long* h, int v, int u) {
  return (h[v] - h[u] + 1004535809) % 1004535809 * ipw2[dep[u]] % 1004535809;
}
signed main() {
  scanf("%d%s", &n, str + 1);
  for (int i = 1, iend = n; i <= iend; ++i) h[i] = str[i] - 'a' + 1;
  for (int i = 2, iend = n; i <= iend; ++i) {
    int x = read(), y = read();
    adde(x, y), adde(y, x);
  }
  lg[0] = -1;
  for (int i = 1, iend = n << 1; i <= iend; ++i) lg[i] = lg[i >> 1] + 1;
  pw1[0] = 1;
  for (int i = 1, iend = n; i <= iend; ++i)
    pw1[i] = 1ll * pw1[i - 1] * 31 % 998244353;
  pw2[0] = 1;
  for (int i = 1, iend = n; i <= iend; ++i)
    pw2[i] = 1ll * pw2[i - 1] * 37 % 1004535809;
  ipw1[n] = qpow(pw1[n], 998244353 - 2, 998244353);
  for (int i = n - 1, iend = 0; i >= iend; --i)
    ipw1[i] = 1ll * ipw1[i + 1] * 31 % 998244353;
  ipw2[n] = qpow(pw2[n], 1004535809 - 2, 1004535809);
  for (int i = n - 1, iend = 0; i >= iend; --i)
    ipw2[i] = 1ll * ipw2[i + 1] * 37 % 1004535809;
  dfs1(1, 0), dfs2(1, 1);
  for (int i = 1, iend = lg[tmr]; i <= iend; ++i)
    for (int j = 1, jend = tmr; j <= jend; ++j) {
      int x = ST[i - 1][j], y = ST[i - 1][j + (1 << (i - 1))];
      ST[i][j] = dep[x] < dep[y] ? x : y;
    }
  for (m = read(); m; --m) {
    int a = read(), b = read(), c = read(), d = read();
    int lca1 = LCA(a, b), lca2 = LCA(c, d);
    int dis1 = dep[a] + dep[b] - 2 * dep[lca1],
        dis2 = dep[c] + dep[d] - 2 * dep[lca2];
    int l = 0, r = min(dis1, dis2), res = -1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      long long H1 =
          dep[a] - dep[lca1] >= mid
              ? calc1(u1, a, kthanc(a, mid))
              : (calc1(u1, a, lca1) + calc3(d1, kthanc(b, dis1 - mid), lca1) *
                                          pw1[dep[a] - dep[lca1] + 1]) %
                    998244353;
      long long H2 =
          dep[c] - dep[lca2] >= mid
              ? calc1(u1, c, kthanc(c, mid))
              : (calc1(u1, c, lca2) + calc3(d1, kthanc(d, dis2 - mid), lca2) *
                                          pw1[dep[c] - dep[lca2] + 1]) %
                    998244353;
      long long H3 =
          dep[a] - dep[lca1] >= mid
              ? calc2(u2, a, kthanc(a, mid))
              : (calc2(u2, a, lca1) + calc4(d2, kthanc(b, dis1 - mid), lca1) *
                                          pw2[dep[a] - dep[lca1] + 1]) %
                    1004535809;
      long long H4 =
          dep[c] - dep[lca2] >= mid
              ? calc2(u2, c, kthanc(c, mid))
              : (calc2(u2, c, lca2) + calc4(d2, kthanc(d, dis2 - mid), lca2) *
                                          pw2[dep[c] - dep[lca2] + 1]) %
                    1004535809;
      if (H1 == H2 && H3 == H4)
        l = mid + 1, res = mid;
      else
        r = mid - 1;
    }
    printf("%d\n", ++res);
  }
}