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;
char In[1 << 20], *ss = In, *tt = In;
long long read() {
  long long x = 0, f = 1;
  char ch =
      (ss == tt && (tt = (ss = In) + fread(In, 1, 1 << 20, stdin), ss == tt)
           ? EOF
           : *ss++);
  for (; ch < '0' || ch > '9';
       ch = (ss == tt &&
                     (tt = (ss = In) + fread(In, 1, 1 << 20, stdin), ss == tt)
                 ? EOF
                 : *ss++))
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9';
       ch = (ss == tt &&
                     (tt = (ss = In) + fread(In, 1, 1 << 20, stdin), ss == tt)
                 ? EOF
                 : *ss++))
    x = x * 10 + int(ch - '0');
  return x * f;
}
char readc() {
  char ch =
      (ss == tt && (tt = (ss = In) + fread(In, 1, 1 << 20, stdin), ss == tt)
           ? EOF
           : *ss++);
  for (; ch < 'a' || ch > 'z';
       ch = (ss == tt &&
                     (tt = (ss = In) + fread(In, 1, 1 << 20, stdin), ss == tt)
                 ? EOF
                 : *ss++))
    ;
  return ch;
}
const int MAXN = 6e5 + 5, MAXM = 1e6 + 5, P = 612321329;
int pls(int a, int b) {
  a += b;
  return a >= P ? a - P : a;
}
int mns(int a, int b) {
  a -= b;
  return a < 0 ? a + P : a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int qpow(int a, int n) {
  int ret = 1;
  for (; n; n >>= 1, a = mul(a, a))
    if (n & 1) ret = mul(ret, a);
  return ret;
}
int n, m, head[MAXN], ver[MAXN], nxt[MAXN], cnt, st[MAXN][20], _st, dep[MAXN],
    pos[MAXN], lg2[MAXN], dfn[MAXN], tim, top[MAXN], son[MAXN], U[MAXN],
    D[MAXN], h[MAXN], fa[MAXN][20];
char ch[MAXN];
struct hsint {
  int a, b;
  hsint(int a = 0, int b = 0) : a(a), b(b) {}
};
const hsint B = hsint(21313, 3137), invB = hsint(488380344, 475100451);
hsint operator+(const hsint& x, const hsint& y) {
  return hsint(pls(x.a, y.a), pls(x.b, y.b));
}
hsint operator-(const hsint& x, const hsint& y) {
  return hsint(mns(x.a, y.a), mns(x.b, y.b));
}
hsint operator*(const hsint& x, const hsint& y) {
  return hsint(mul(x.a, y.a), mul(x.b, y.b));
}
bool operator==(const hsint& x, const hsint& y) {
  return x.a == y.a && x.b == y.b;
}
hsint pre[MAXN], suf[MAXN], powB[MAXN], powinvB[MAXN];
void addedge(int u, int v) {
  ver[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
}
namespace LCA {
void dfs1(int u, int f) {
  st[++_st][0] = u;
  pos[u] = _st;
  dep[u] = dep[f] + 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    dfs1(v, u);
    st[++_st][0] = u;
  }
}
int better(int u, int v) { return dep[u] < dep[v] ? u : v; }
void init_st() {
  for (int k = 1; (1 << k) <= _st; k++)
    for (int i = 1; i + (1 << k) - 1 <= _st; i++)
      st[i][k] = better(st[i][k - 1], st[i + (1 << (k - 1))][k - 1]);
  lg2[0] = -1;
  for (int i = 1; i <= _st; i++) lg2[i] = lg2[i >> 1] + 1;
}
int Lca(int u, int v) {
  u = pos[u], v = pos[v];
  if (u > v) swap(u, v);
  int k = lg2[v - u + 1];
  return better(st[u][k], st[v - (1 << k) + 1][k]);
}
int Dist(int u, int v) { return dep[u] + dep[v] - 2 * dep[Lca(u, v)]; }
void init_LCA() {
  dfs1(1, 0);
  init_st();
}
}  // namespace LCA
using namespace LCA;
namespace KTH {
void dfs2(int u, int f) {
  fa[u][0] = f;
  h[u] = 1;
  for (int k = 1; k <= 19; k++) fa[u][k] = fa[fa[u][k - 1]][k - 1];
  for (int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    dfs2(v, u);
    h[u] = max(h[u], h[v] + 1);
    if (h[v] > h[son[u]]) son[u] = v;
  }
}
void dfs3(int u, int tp, int up) {
  dfn[u] = ++tim;
  U[tim] = up;
  D[tim] = u;
  top[u] = tp;
  if (son[u]) dfs3(son[u], tp, fa[up][0]);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == fa[u][0] || v == son[u]) continue;
    dfs3(v, v, v);
  }
}
int Kth(int u, int k) {
  if (!k) return u;
  u = fa[u][lg2[k]];
  k -= (1 << lg2[k]);
  k -= dep[u] - dep[top[u]];
  u = top[u];
  if (k >= 0)
    return U[dfn[u] + k];
  else
    return D[dfn[u] - k];
}
void init_Kth() {
  dfs2(1, 0);
  dfs3(1, 1, 1);
}
}  // namespace KTH
using namespace KTH;
hsint pre_get(int u, int v) {
  return pre[v] - pre[fa[u][0]] * powB[dep[v] - dep[u] + 1];
}
hsint suf_get(int u, int v) {
  return (suf[u] - suf[fa[v][0]]) * powinvB[dep[v] - 1];
}
hsint geths(int u, int v, int len) {
  int p = Lca(u, v);
  if (len == 0)
    return 0;
  else if (len <= dep[u] - dep[p] + 1) {
    return suf_get(u, Kth(u, len - 1));
  } else {
    hsint p1 =
        dep[u] - dep[p] - 1 >= 0 ? suf_get(u, Kth(u, dep[u] - dep[p] - 1)) : 0;
    int lst = len - (dep[u] - dep[p]), k = dep[v] - dep[p] + 1;
    hsint p2 = pre_get(p, Kth(v, k - lst));
    return p1 * powB[lst] + p2;
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) ch[i] = readc();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    addedge(u, v);
    addedge(v, u);
  }
  init_LCA();
  init_Kth();
  powB[0] = powinvB[0] = hsint(1, 1);
  for (int i = 1; i <= n; i++) {
    powB[i] = powB[i - 1] * B;
    powinvB[i] = powinvB[i - 1] * invB;
  }
  for (int i = 1; i <= n; i++) {
    int u = D[i], f = fa[u][0];
    pre[u] = pre[f] * B + hsint(ch[u], ch[u]);
    suf[u] = suf[f] + hsint(ch[u], ch[u]) * powB[dep[u] - 1];
  }
  for (int te = read(); te; te--) {
    int a = read(), b = read(), c = read(), d = read();
    int l = 0, r = min(Dist(a, b) + 1, Dist(c, d) + 1), ans = -1;
    while (l <= r) {
      int m = (l + r) >> 1;
      if (geths(a, b, m) == geths(c, d, m))
        ans = m, l = m + 1;
      else
        r = m - 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}