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;
const int inf = int(1e9);
const double eps = 1e-4;
const double pi = 4 * atan(double(1));
const int N = int(3e5) + 100;
const int LOG = 19;
const int M = int(1e6) + 100;
const int X1 = 4243;
const int X2 = 10007;
const int MOD1 = int(1e9) + 7;
const int MOD2 = 42424243;
struct _hash {
  int a;
  _hash() {}
  _hash(int a) : a(a) {}
};
_hash operator+(const _hash &a, const _hash &b) {
  _hash c(a.a + b.a);
  if (c.a >= MOD1) {
    c.a -= MOD1;
  }
  return c;
}
_hash operator-(const _hash &a, const _hash &b) {
  _hash c(a.a - b.a);
  if (c.a < 0) {
    c.a += MOD1;
  }
  return c;
}
_hash operator*(const _hash &a, const _hash &b) {
  return _hash((1LL * a.a * b.a) % MOD1);
}
bool operator==(const _hash &a, const _hash &b) { return a.a == b.a; }
const _hash X(X1);
struct tq {
  int a, b, ab, c, d, cd;
};
struct query {
  int pv, pu, u, num;
  query(int pv, int pu, int u, int num) : pv(pv), pu(pu), u(u), num(num) {}
};
int n, sz;
bool used[N];
char s[N];
int _prev[N], _rank[N], val[N], h[N], st[N];
int p[LOG][N];
tq q[M];
int ans[M];
vector<int> g[N];
vector<pair<int, int *> > lca[N];
vector<query> queries[N];
_hash sum[N], sum_rev[N], pw[2 * N];
inline void add(int a, int b, int *ab) {
  lca[a].push_back(make_pair(b, ab));
  lca[b].push_back(make_pair(a, ab));
}
int get(int a) {
  if (_prev[a] != a) {
    _prev[a] = get(_prev[a]);
  }
  return _prev[a];
}
inline void merge(int a, int b, int c) {
  a = get(a);
  b = get(b);
  if (_rank[a] < _rank[b]) {
    swap(a, b);
  }
  _prev[b] = a;
  if (_rank[a] == _rank[b]) {
    ++_rank[a];
  }
  val[a] = c;
}
void dfs(int v, int pv = -1) {
  p[0][v] = pv;
  used[v] = true;
  if (pv != -1) {
    g[v].erase(find((g[v]).begin(), (g[v]).end(), pv));
    h[v] = h[pv] + 1;
  } else {
    h[v] = 0;
  }
  sum[v] = (pv == -1 ? _hash(0) : sum[pv]) + pw[h[v]] * _hash(s[v]);
  sum_rev[v] = (pv == -1 ? _hash(0) : sum_rev[pv]) + pw[n - h[v]] * _hash(s[v]);
  for (int i = 0; i < int((lca[v]).size()); ++i) {
    if (used[lca[v][i].first]) {
      *(lca[v][i].second) = val[get(lca[v][i].first)];
    }
  }
  for (int i = 0; i < int((g[v]).size()); ++i) {
    dfs(g[v][i], v);
    merge(g[v][i], v, v);
  }
}
inline int up(int v, int need) {
  if (need == 0) {
    return v;
  }
  for (int i = LOG - 1; i >= 0; --i) {
    if (need >= (1 << i)) {
      v = p[i][v];
      need -= (1 << i);
    }
  }
  return v;
}
inline _hash get_sum_rev(int v, int pv) {
  return (sum_rev[v] - (p[0][pv] == -1 ? _hash(0) : sum_rev[p[0][pv]])) *
         pw[h[v]];
}
inline int up(int v, int u, int pv) {
  int res = 0;
  for (int i = LOG - 1; i >= 0; --i) {
    if (p[i][v] != -1 && h[p[i][v]] >= h[pv]) {
      int nv = p[i][v];
      int nu = p[i][u];
      if (get_sum_rev(v, nv) == get_sum_rev(u, nu)) {
        res += (1 << i);
        v = nv;
        u = nu;
      }
    }
  }
  if (s[v] == s[u]) {
    ++res;
  }
  return res;
}
inline _hash get_sum(int v, int pv) {
  return (sum[v] - (p[0][pv] == -1 ? _hash(0) : sum[p[0][pv]])) * pw[n - h[pv]];
}
inline int down(int pv, int v, int pu, int u) {
  if (h[v] - h[pv] > h[u] - h[pu]) {
    swap(v, u);
    swap(pv, pu);
  }
  u = up(u, (h[u] - h[pu]) - (h[v] - h[pv]));
  int res = 0;
  for (int i = LOG - 1; i >= 0; --i) {
    if (p[i][v] != -1 && h[p[i][v]] > h[pv]) {
      int nv = p[i][v];
      int nu = p[i][u];
      if (get_sum_rev(nv, pv) == get_sum_rev(nu, pu)) {
        res += (h[v] - h[pv]) - (1 << i);
        pv = nv;
        pu = nu;
      } else {
        v = nv;
        u = nu;
      }
    }
  }
  if (pv != v && s[v] == s[u]) {
    ++res;
  }
  return res;
}
inline void process(int a, int b, int ab, int c, int d, int cd, int num) {
  if (h[a] - h[ab] > h[c] - h[cd]) {
    swap(a, c);
    swap(b, d);
    swap(ab, cd);
  }
  int tmp = up(c, h[a] - h[ab]);
  if (get_sum_rev(a, ab) == get_sum_rev(c, tmp)) {
    if (tmp == cd) {
      ans[num] = (h[a] - h[ab] + 1) + down(ab, b, cd, d);
    } else {
      int len = min(h[b] - h[ab], h[tmp] - h[cd]);
      int tmp2 = up(b, (h[b] - h[ab]) - len);
      int tmp3 = up(tmp, len);
      if (get_sum(tmp2, ab) == get_sum_rev(tmp, tmp3)) {
        ans[num] = (h[a] - h[ab] + 1) + (h[tmp2] - h[ab]);
        if (tmp3 == cd) {
          ans[num] += down(tmp2, b, cd, d);
        }
      } else {
        ans[num] = (h[a] - h[ab] + 1);
        queries[tmp2].push_back(query(ab, tmp3, tmp, num));
      }
    }
  } else {
    ans[num] = up(a, c, ab);
  }
}
int cnt = 0;
void dfs2(int v) {
  ++cnt;
  st[sz++] = v;
  for (int i = 0; i < int((queries[v]).size()); ++i) {
    int pv = queries[v][i].pv, pu = queries[v][i].pu, u = queries[v][i].u,
        num = queries[v][i].num;
    for (int j = LOG - 1; j >= 0; --j) {
      if (p[j][u] != -1 && h[p[j][u]] >= h[pu]) {
        int nu = p[j][u];
        int nv = st[sz - 1 - (h[v] - h[pv]) + (1 << j)];
        if (get_sum_rev(u, nu) == get_sum(nv, pv)) {
          ans[num] += (1 << j);
          u = nu;
          pv = nv;
        }
      }
    }
  }
  for (int i = 0; i < int((g[v]).size()); ++i) {
    dfs2(g[v][i]);
  }
  --sz;
}
int main() {
  scanf("%d", &n);
  gets(s);
  gets(s);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d %d %d", &q[i].a, &q[i].b, &q[i].c, &q[i].d);
    --q[i].a;
    --q[i].b;
    --q[i].c;
    --q[i].d;
    add(q[i].a, q[i].b, &q[i].ab);
    add(q[i].c, q[i].d, &q[i].cd);
  }
  for (int i = 0; i < n; ++i) {
    _prev[i] = i;
    _rank[i] = 1;
    val[i] = i;
  }
  pw[0] = _hash(1);
  for (int i = 1; i <= 2 * n; ++i) {
    pw[i] = pw[i - 1] * X;
  }
  dfs(0);
  for (int i = 1; i < LOG; ++i) {
    for (int j = 0; j < n; ++j) {
      if (p[i - 1][j] == -1) {
        p[i][j] = -1;
        continue;
      }
      p[i][j] = p[i - 1][p[i - 1][j]];
    }
  }
  for (int i = 0; i < m; ++i) {
    process(q[i].a, q[i].b, q[i].ab, q[i].c, q[i].d, q[i].cd, i);
  }
  sz = 0;
  dfs2(0);
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}