Problem: As we all know, Eleven has special abilities. Thus, Hopper convinced her to close the gate to the Upside Down World with her mind. Upside down monsters like to move between the worlds, so they are going to attack Hopper and Eleven in order to make them stop. The monsters live in the vines. The vines form a tree with n vertices, numbered from 1 through n. There's a lowercase English letter written in each tunnel (edge).

<image>

Upside down is a magical world. There are m types of monsters in upside down, numbered from 1 through m. Each type of monster has a special word that gives them powers. The special word of type i is si. There are q monsters in upside down. Each one is at a junction (vertex) and is going to some other junction. If monster of type k goes from junction i to junction j, the power it gains is the number of times it sees its special world (sk) consecutively in the tunnels. More formally: 

If f(i, j) is the string we get when we concatenate the letters written in the tunnels on the shortest path from i to j, then the power the monster gains is the number of occurrences of sk in f(i, j).

Hopper and Eleven want to get prepared, so for each monster, they want to know the power the monster gains after moving. 

Input

The first line of input contains three integers, n, m and q (2 ≤ n ≤ 105, 1 ≤ m, q ≤ 105).

The next n - 1 lines contain the tunnels (edges). Each line contains two integers v and u (1 ≤ v, u ≤ n, v ≠ u) and a lowercase English letter c, meaning there's a tunnel connecting junctions v and u written c in it. It is guaranteed that the given graph is a tree.

The next m lines contain the special words. i-th line of them contains a single string si (1 ≤ |si| ≤ 105), consisting of lowercase English letters. It is guaranteed that |s1| + |s2| + ... + |sm| ≤ 105).

The next q lines contain the monsters. Each line contains three integers i, j and k (1 ≤ i, j ≤ n, i ≠ j, 1 ≤ k ≤ m), meaning a monster of type k is going from junction number i to junction number j.

Output

Print q lines. i-th line should contain a single integer, the power the i-th monster gains after moving.

Examples

Input

6 4 5
1 6 b
2 3 a
1 2 b
5 3 b
4 5 b
a
b
bb
aa
1 2 1
6 2 3
1 6 2
4 5 4
1 6 2


Output

0
1
1
0
1


Input

10 6 7
1 3 s
10 1 d
2 6 s
5 2 d
7 4 l
8 9 d
8 10 l
7 2 d
8 7 l
dl
dssld
d
d
l
sl
4 5 4
3 7 5
10 6 2
3 1 4
7 5 6
10 9 4
9 8 4


Output

2
2
0
0
0
1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int tree[300005 * 20], lefto[300005 * 20], righto[300005 * 20], nw;
struct ext {
  int len, link;
  map<char, int> next;
};
ext st[300005];
int sz, last;
void sa_init() {
  sz = last = 0;
  st[0].len = 0;
  st[0].link = -1;
  ++sz;
}
int update(int node, int st, int en, int g) {
  int ps = ++nw;
  if (st == en) {
    tree[ps] = tree[node] + 1;
    return ps;
  }
  lefto[ps] = lefto[node];
  righto[ps] = righto[node];
  int mid = (st + en) >> 1;
  if (g <= mid)
    lefto[ps] = update(lefto[node], st, mid, g);
  else
    righto[ps] = update(righto[node], mid + 1, en, g);
  tree[ps] = tree[lefto[ps]] + tree[righto[ps]];
  return ps;
}
int query(int node, int node2, int st, int en, int l, int r) {
  if (st >= l and en <= r) return tree[node] - tree[node2];
  int mid = (st + en) >> 1;
  int ret = 0;
  if (l <= mid) ret += query(lefto[node], lefto[node2], st, mid, l, r);
  if (r > mid) ret += query(righto[node], righto[node2], mid + 1, en, l, r);
  return ret;
}
int dp[300005][26];
int extend(char c, int last) {
  if (dp[last][c - 97] != -1) return dp[last][c - 97];
  int cur = sz++;
  st[cur].len = st[last].len + 1;
  int p;
  for (p = last; p != -1 && !st[p].next.count(c); p = st[p].link)
    st[p].next[c] = cur;
  if (p == -1)
    st[cur].link = 0;
  else {
    int q = st[p].next[c];
    if (st[p].len + 1 == st[q].len)
      st[cur].link = q;
    else {
      int clone = sz++;
      st[clone].len = st[p].len + 1;
      st[clone].next = st[q].next;
      st[clone].link = st[q].link;
      for (; p != -1 && st[p].next[c] == q; p = st[p].link)
        st[p].next[c] = clone;
      st[q].link = st[cur].link = clone;
    }
  }
  return dp[last][c - 97] = cur;
}
struct st2 {
  int g, lev, id, bol;
  st2() {}
  st2(int _g, int _lev, int _id, int _bol) {
    g = _g;
    lev = _lev;
    id = _id;
    bol = _bol;
  }
};
int top[300005], P[300005];
vector<st2> danda[300005];
void upchain(int lev, int node, int c) {
  while (node) {
    if (st[node].len == 0) return;
    int g = lev - st[node].len;
    assert(g >= 1);
    if (c == -1) g = 1e8;
    danda[top[node]].push_back(st2(g, lev, -1, -1));
    node = P[top[node]];
  }
}
int mini(int v) {
  if (v == 0) return 0;
  return st[st[v].link].len + 1;
}
void query_chain(int node, int lev, int lc, int q, int bol, int len) {
  while (node) {
    int k = min(st[node].len, len);
    if (k == 0) return;
    int g = min(lc + k, lev);
    danda[top[node]].push_back(st2(lc, g, q, bol));
    node = P[top[node]];
  }
}
int pos, start[300005], endo[300005];
struct st1 {
  int node, lev, id;
  st1() {}
  st1(int _node, int _lev, int _id) {
    node = _node;
    lev = _lev;
    id = _id;
  }
};
vector<st1> ques[300005];
struct st3 {
  int id, bol, node, lc;
  st3() {}
  st3(int _node, int _lc, int _id, int _bol) {
    node = _node;
    lc = _lc;
    id = _id;
    bol = _bol;
  }
};
vector<st3> half[300005];
vector<int> liqu[300005];
int root[300005], ans[300005];
vector<int> vc[300005];
vector<char> cost[300005];
void dfs(int v, int par, int prev, char c, int lev) {
  root[lev] = root[lev - 1];
  if (v > 1) {
    prev = st[prev].next[c];
    assert(prev != 0);
    root[lev] = update(root[lev], 1, pos, start[prev]);
    upchain(lev, prev, 1);
  }
  for (int i = 0; i < ques[v].size(); i++) {
    int nd = ques[v][i].node;
    int le = ques[v][i].lev;
    int id = ques[v][i].id;
    int g = query(root[lev], root[le - 1], 1, pos, start[nd], endo[nd]);
    ans[id] += g;
  }
  for (int i = 0; i < half[v].size(); i++) {
    int id = half[v][i].id;
    int bol = half[v][i].bol;
    int node = half[v][i].node;
    int lc = half[v][i].lc;
    query_chain(node, lev, lc, id, bol, liqu[v][i]);
  }
  for (int i = 0; i < vc[v].size(); i++) {
    int w = vc[v][i];
    if (w == par) continue;
    dfs(w, v, prev, cost[v][i], lev + 1);
  }
  if (v > 1) {
    upchain(lev, prev, -1);
  }
}
int bap[300005][20], L[300005], len;
void dfs2(int v, int par, char c, int last, int lev) {
  bap[v][0] = par;
  L[v] = lev;
  len = max(len, lev);
  if (v > 1) {
    last = extend(c, last);
  }
  for (int i = 0; i < vc[v].size(); i++) {
    int w = vc[v][i];
    if (w == par) continue;
    dfs2(w, v, cost[v][i], last, lev + 1);
  }
}
int maxi[300005], siz[300005];
vector<int> ac[300005];
void dfs3(int v) {
  start[v] = ++pos;
  maxi[v] = -1;
  for (int i = 0; i < ac[v].size(); i++) {
    int w = ac[v][i];
    dfs3(w);
    if (maxi[v] == -1)
      maxi[v] = w;
    else if (siz[maxi[v]] < siz[w]) {
      maxi[v] = w;
    }
    siz[v] += siz[w];
  }
  endo[v] = pos;
  siz[v]++;
}
void hld(int v, int par, int r) {
  if (r == -1)
    top[v] = v;
  else
    top[v] = r;
  P[v] = par;
  if (maxi[v] != -1) hld(maxi[v], v, top[v]);
  for (int i = 0; i < ac[v].size(); i++) {
    int w = ac[v][i];
    if (w == maxi[v]) continue;
    hld(w, v, -1);
  }
}
int tree2[300005 * 20];
void update2(int node, int st, int en, int l, int g) {
  if (st == en) {
    tree2[node] = g;
    return;
  }
  int mid = (st + en) >> 1;
  if (l <= mid)
    update2(node * 2, st, mid, l, g);
  else
    update2(node * 2 + 1, mid + 1, en, l, g);
  tree2[node] = min(tree2[node * 2], tree2[node * 2 + 1]);
}
int query2(int node, int st, int en, int g, int l) {
  if (tree2[node] > g) return -1;
  if (st > l) return -1;
  if (st == en) {
    return st;
  }
  int mid = (st + en) >> 1;
  int r = query2(node * 2 + 1, mid + 1, en, g, l);
  if (r > -1) return r;
  return query2(node * 2, st, mid, g, l);
}
int create[300005][2];
void sim(st2 ps) {
  int lev = ps.lev;
  int g = ps.g;
  int ans = query2(1, 1, len, g, lev);
  if (ans > lev || ans <= g) return;
  int id = ps.id;
  int bol = ps.bol;
  create[id][bol] = max(create[id][bol], ans - g);
}
void simulation(int n) {
  for (int i = 0; i < n; i++) {
    if (danda[i].size() == 0) continue;
    for (int j = 0; j < danda[i].size(); j++) {
      if (danda[i][j].bol >= 0) {
        sim(danda[i][j]);
        continue;
      }
      int lev = danda[i][j].lev;
      int g = danda[i][j].g;
      update2(1, 1, len, lev, g);
    }
  }
}
int beg[300005][2], en[300005][2], obo[2];
vector<int> trie[2][300005];
void dfs5(int v, int bol) {
  beg[v][bol] = ++obo[bol];
  for (int i = 0; i < trie[bol][v].size(); i++) {
    int w = trie[bol][v][i];
    dfs5(w, bol);
  }
  en[v][bol] = obo[bol];
}
vector<int> sub[300005];
int update3(int node, int st, int en, int l, int r) {
  int ps = ++nw;
  assert(nw < 300005 * 20);
  tree[ps] = tree[node];
  lefto[ps] = lefto[node];
  righto[ps] = righto[node];
  if (st >= l and en <= r) {
    tree[ps] += 1;
    return ps;
  }
  int mid = (st + en) >> 1;
  if (mid >= l) lefto[ps] = update3(lefto[node], st, mid, l, r);
  if (mid < r) righto[ps] = update3(righto[node], mid + 1, en, l, r);
  return ps;
}
int vis2[300005], casio = 1;
void run_sim(int v, int par) {
  root[v] = root[par];
  for (int i = 0; i < sub[v].size(); i++) {
    int w = sub[v][i];
    root[v] = update3(root[v], 1, obo[1], beg[w][1], en[w][1]);
  }
  for (int i = 0; i < trie[0][v].size(); i++) {
    int w = trie[0][v][i];
    if (vis2[w] != casio) continue;
    run_sim(w, v);
  }
}
int query3(int node, int st, int en, int l) {
  int ret = tree[node];
  if (st == en) return ret;
  int mid = (st + en) >> 1;
  if (l <= mid)
    ret += query3(lefto[node], st, mid, l);
  else
    ret += query3(righto[node], mid + 1, en, l);
  return ret;
}
map<pair<pair<int, int>, int>, int> name[2];
int cnt[2];
int prime[] = {103, 10007, 37};
int mod = 784568360, mod2 = 820925358, mod3 = 1e9 + 7;
int crt(int sum, int sum2, int sum3, int bol) {
  if (name[bol][make_pair(make_pair(sum, sum2), sum3)] == 0) {
    name[bol][make_pair(make_pair(sum, sum2), sum3)] = ++cnt[bol];
  }
  return name[bol][make_pair(make_pair(sum, sum2), sum3)];
}
int lps[300005];
vector<int> linear[2][300005];
int vis[300005][2];
void func(string pat, int bol, int id) {
  int M = pat.size();
  linear[bol][id].resize(M + 2);
  long long sum = 0, sum2 = 0, sum3 = 0;
  for (int i = 0; i < M; i++) {
    sum = sum * prime[0] + pat[i] - 96;
    sum %= mod;
    sum2 = sum2 * prime[1] + pat[i] - 96;
    sum2 %= mod2;
    sum3 = sum3 * prime[2] + pat[i] - 96;
    sum3 %= mod3;
    int g = crt(sum, sum2, sum3, bol);
    linear[bol][id][i + 1] = g;
  }
  int len = 0;
  lps[0] = 0;
  int i = 1;
  while (i < M) {
    if (pat[i] == pat[len]) {
      len++;
      lps[i] = len;
      i++;
    } else {
      if (len != 0) {
        len = lps[len - 1];
      } else {
        lps[i] = 0;
        i++;
      }
    }
  }
  for (int i = 0; i < M; i++) {
    int r = linear[bol][id][i + 1];
    if (vis[r][bol]) continue;
    vis[r][bol] = 1;
    int g = lps[i];
    int x = linear[bol][id][g];
    trie[bol][x].push_back(r);
  }
}
void init(int n) {
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n; i++)
      if (bap[i][j - 1] != -1) bap[i][j] = bap[bap[i][j - 1]][j - 1];
}
int lca_query(int p, int q) {
  int tmp, log, i;
  if (L[p] < L[q]) tmp = p, p = q, q = tmp;
  log = 1;
  while (1) {
    int next = log + 1;
    if ((1 << next) > L[p]) break;
    log++;
  }
  for (i = log; i >= 0; i--)
    if (L[p] - (1 << i) >= L[q]) p = bap[p][i];
  if (p == q) return p;
  for (i = log; i >= 0; i--)
    if (bap[p][i] != -1 && bap[p][i] != bap[q][i]) p = bap[p][i], q = bap[q][i];
  return bap[p][0];
}
string my[300005];
char str2[300005];
int hf[300005][2], dir[300005][2], lin[300005][2];
int qui[300005];
map<string, int> coun;
int ara[300005];
vector<int> temp[300005];
int main() {
  int i, j, k, l, m, n, q, ff;
  for (int i = 1; i < 300005 * 20; i++) tree2[i] = 1e8;
  memset(dp, -1, sizeof dp);
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%s", &l, &k, str2);
    vc[l].push_back(k);
    vc[k].push_back(l);
    cost[l].push_back(str2[0]);
    cost[k].push_back(str2[0]);
  }
  sa_init();
  memset(bap, -1, sizeof bap);
  dfs2(1, -1, 0, 0, 1);
  for (int kk = 1; kk <= m; kk++) {
    scanf("%s", str2);
    my[kk] += str2;
    int l = strlen(str2);
    int cur = 0;
    int ck = 1, len = 0;
    for (int i = l - 1; i >= 0; i--) {
      char c = str2[i];
      int prv = len;
      if (st[cur].next.find(c) == st[cur].next.end())
        ck = 0;
      else {
        len++;
        cur = st[cur].next[c];
        continue;
      }
      while (cur > -1 && st[cur].next.find(c) == st[cur].next.end()) {
        cur = st[cur].link;
      }
      if (cur == -1) {
        len = 0;
        cur = 0;
        continue;
      }
      len = min(st[cur].len, prv);
      len++;
      cur = st[cur].next[c];
    }
    hf[kk][0] = cur;
    lin[kk][0] = len;
    if (ck == 0) cur = 0;
    dir[kk][0] = cur;
    cur = 0, ck = 1, len = 0;
    for (int i = 0; str2[i]; i++) {
      char c = str2[i];
      int prv = len;
      if (st[cur].next.find(c) == st[cur].next.end())
        ck = 0;
      else {
        len++;
        cur = st[cur].next[c];
        continue;
      }
      while (cur > -1 && st[cur].next.find(c) == st[cur].next.end()) {
        cur = st[cur].link;
      }
      if (cur == -1) {
        len = 0;
        cur = 0;
        continue;
      }
      len = min(st[cur].len, prv);
      len++;
      cur = st[cur].next[c];
    }
    hf[kk][1] = cur;
    lin[kk][1] = len;
    if (ck == 0) cur = 0;
    dir[kk][1] = cur;
  }
  init(n);
  for (int i = 1; i < sz; i++) {
    ac[st[i].link].push_back(i);
  }
  dfs3(0);
  hld(0, 0, -1);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &l, &k, &j);
    qui[i] = j;
    int lc = lca_query(l, k);
    if (l == lc) {
      int x = L[lc] + my[j].size();
      if (dir[j][1] && x <= L[k]) ques[k].push_back(st1(dir[j][1], x, i));
      continue;
    }
    if (k == lc) {
      int x = L[lc] + my[j].size();
      if (dir[j][0] && x <= L[l]) ques[l].push_back(st1(dir[j][0], x, i));
      continue;
    }
    int x = L[lc] + my[j].size();
    if (dir[j][0] && x <= L[l]) ques[l].push_back(st1(dir[j][0], x, i));
    if (dir[j][1] && x <= L[k]) {
      ques[k].push_back(st1(dir[j][1], x, i));
    }
    if (hf[j][0])
      half[l].push_back(st3(hf[j][0], L[lc], i, 0)),
          liqu[l].push_back(lin[j][0]);
    if (hf[j][1])
      half[k].push_back(st3(hf[j][1], L[lc], i, 1)),
          liqu[k].push_back(lin[j][1]);
  }
  dfs(1, -1, 0, 0, 1);
  simulation(sz);
  for (int i = 1; i <= m; i++) {
    long long sum = 0, sum2 = 0;
    func(my[i], 0, i);
    reverse(my[i].begin(), my[i].end());
    func(my[i], 1, i);
  }
  for (int i = 0; i <= nw; i++) {
    tree[i] = 0;
    lefto[i] = 0;
    righto[i] = 0;
  }
  nw = 0;
  dfs5(0, 0);
  dfs5(0, 1);
  for (int i = 1; i <= q; i++) {
    if (create[i][0] == 0 || create[i][1] == 0) continue;
    j = qui[i];
    temp[j].push_back(i);
  }
  for (int i = 1; i <= m; i++) {
    l = my[i].size();
    casio++;
    vector<int> tp;
    for (int j = 1; j < l; j++) {
      int g = linear[0][i][j];
      tp.push_back(g);
      int h = linear[1][i][l - j];
      vis2[g] = casio;
      sub[g].push_back(h);
    }
    int g = linear[0][i][l];
    vis2[g] = casio;
    tp.push_back(g);
    run_sim(0, 0);
    for (j = 0; j < temp[i].size(); j++) {
      int q = temp[i][j];
      int g = linear[0][i][create[q][0]];
      int h = linear[1][i][create[q][1]];
      ans[q] += query3(root[g], 1, obo[1], beg[h][1]);
    }
    for (int j = 0; j < tp.size(); j++) {
      int w = tp[j];
      sub[w].clear();
    }
    for (int j = 0; j <= nw; j++) {
      tree[j] = 0;
      lefto[j] = 0;
      righto[j] = 0;
    }
    nw = 0;
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
}