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 a14;
inline int rd(int l, int r) { return rand() % (r - l + 1) + l; }
const int mxn = 2e5 + 3;
int n, m, qn, to[mxn], nxt[mxn], fir[mxn], gn = 1, co[mxn];
inline void gadd(int x, int y, char c) {
  to[++gn] = y, nxt[gn] = fir[x], fir[x] = gn, co[gn] = c;
}
int lp[mxn], rp[mxn], bs[mxn], sb[mxn];
char ss[mxn];
int a1[mxn], a2[mxn], a3[mxn], a4[mxn], a5[mxn];
int dfn[mxn], dfa[mxn], siz[mxn], top[mxn], zs[mxn], dep[mxn], fa[mxn], dt,
    qa[mxn];
vector<int> jp[mxn];
inline void dfs1(int x, int f) {
  fa[x] = f, dep[x] = dep[f] + 1, siz[x] = 1;
  for (register int i = fir[x]; i; i = nxt[i])
    if (to[i] != f) {
      dfs1(to[i], x), siz[x] += siz[to[i]];
      if (siz[to[i]] > siz[zs[x]]) zs[x] = to[i];
    }
}
inline void dfs2(int x, int f) {
  top[x] = x == zs[f] ? top[f] : x, dfn[x] = ++dt, dfa[dt] = x;
  jp[top[x]].push_back(x);
  if (zs[x]) dfs2(zs[x], x);
  for (register int i = fir[x]; i; i = nxt[i])
    if (!dfn[to[i]]) dfs2(to[i], x);
}
inline int lca(int x, int y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  return dep[x] < dep[y] ? x : y;
}
inline int kzx(int x, int k) {
  while (x && dep[x] - dep[top[x]] < k)
    k -= dep[x] - dep[top[x]] + 1, x = fa[top[x]];
  if (!x || !k) return x;
  return jp[top[x]][dep[x] - dep[top[x]] - k];
}
struct thr {
  int a, b, c;
};
struct awe {
  int fail[mxn], ac[mxn][26], tt, q[mxn], qh, qt, po[mxn], b1[mxn], b2[mxn], bn;
  vector<thr> vc[mxn];
  inline void w1() {
    for (int i = 1; i <= m; ++i) {
      int d = 0;
      for (int j = rp[i]; j >= lp[i]; --j) {
        int c = ss[j] - 'a';
        if (!ac[d][c]) ac[d][c] = ++tt;
        d = ac[d][c];
      }
      po[i] = d;
    }
    W();
  }
  inline void w2() {
    for (int i = 1; i <= m; ++i) {
      int d = 0;
      for (int j = lp[i]; j <= rp[i]; ++j) {
        int c = ss[j] - 'a';
        if (!ac[d][c]) ac[d][c] = ++tt;
        d = ac[d][c];
      }
      po[i] = d;
    }
    W();
  }
  inline void W() {
    qh = 1, qt = 0;
    for (int c = 0; c < 26; ++c)
      if (ac[0][c]) q[++qt] = ac[0][c];
    while (qh <= qt) {
      int x = q[qh++];
      for (int c = 0; c < 26; ++c)
        if (ac[x][c])
          q[++qt] = ac[x][c], fail[ac[x][c]] = ac[fail[x]][c];
        else
          ac[x][c] = ac[fail[x]][c];
    }
    for (int i = 1; i <= tt; ++i) G(fail[i], i);
    dfs0(0, 0);
    dfs1(1, 0);
  }
  int TO[mxn], NXT[mxn], FIR[mxn], GN;
  inline void G(int x, int y) { TO[++GN] = y, NXT[GN] = FIR[x], FIR[x] = GN; }
  inline void dfs0(int x, int f) {
    b1[x] = ++bn;
    for (register int i = FIR[x]; i; i = NXT[i])
      if (TO[i] != f) dfs0(TO[i], x);
    b2[x] = bn;
  }
  inline void dfs1(int x, int f, int d = 0) {
    add(b1[d], 1);
    for (auto &i : vc[x]) {
      int t = po[i.a];
      qa[i.c] += i.b * ask(b1[t], b2[t]);
    }
    for (register int i = fir[x]; i; i = nxt[i])
      if (to[i] != f) dfs1(to[i], x, ac[d][co[i] - 'a']);
    add(b1[d], -1);
  }
  int ar[mxn];
  inline void add(int x, int y) {
    for (; x <= tt + 1; x += x & -x) ar[x] += y;
  }
  inline int ask(int x) {
    int res = 0;
    for (; x; x -= x & -x) res += ar[x];
    return res;
  }
  inline int ask(int l, int r) { return ask(r) - ask(l - 1); }
} A1, A2;
const int B = 137, mod = 1e9 + 9;
long long bin[mxn];
int pp[mxn][2];
struct aweg {
  int sa[mxn], rk[mxn], tmp[mxn], cnt[mxn], sn;
  char s[mxn], s1[mxn];
  long long ha[mxn], hs[mxn];
  vector<thr> oh[mxn];
  bool rv;
  inline long long ha1(int l, int r) {
    return (ha[r] - ha[l - 1] * bin[r - l + 1] % mod + mod) % mod;
  }
  inline long long ha2(int l, int r) {
    return (hs[r] - hs[l - 1] * bin[r - l + 1] % mod + mod) % mod;
  }
  int bor[mxn], ju[mxn], der[mxn];
  inline void SA(int n, char s[], int rk[], int sa[]) {
    for (int i = 1; i <= n; ++i) rk[i] = s[i], ++cnt[rk[i]];
    for (int i = 1; i <= 200; ++i) cnt[i] += cnt[i - 1];
    for (int i = n; i; --i) sa[cnt[rk[i]]--] = i;
    for (int i = 0; i <= 200; ++i) cnt[i] = 0;
    for (int i = 1; i <= 2 * n; ++i) tmp[i] = 0;
    for (int k = 1, m = 200;; k *= 2) {
      int p = 0;
      for (int i = n - k + 1; i <= n; ++i) tmp[++p] = i;
      for (int i = 1; i <= n; ++i)
        if (sa[i] > k) tmp[++p] = sa[i] - k;
      for (int i = 1; i <= n; ++i) ++cnt[rk[i]];
      for (int i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
      for (int i = n; i; --i) sa[cnt[rk[tmp[i]]]--] = tmp[i];
      for (int i = 1; i <= m; ++i) cnt[i] = 0;
      memcpy(tmp + 1, rk + 1, sizeof(int) * n);
      p = 1, rk[sa[1]] = 1;
      for (int i = 2; i <= n; ++i)
        rk[sa[i]] =
            tmp[sa[i]] == tmp[sa[i - 1]] && tmp[sa[i] + k] == tmp[sa[i - 1] + k]
                ? p
                : ++p;
      if (p == n) break;
      m = p;
    }
  }
  inline void W(bool RV) {
    rv = RV;
    memcpy(s, ss, sizeof(ss));
    if (rv)
      for (int i = 1; i <= m; ++i) reverse(s + lp[i], s + rp[i] + 1);
    sn = rp[m];
    for (int k = 1; k <= m; ++k) {
      int L = lp[k], R = rp[k];
      bor[R + 1] = R + 2;
      for (int i = R, j = R + 2; i >= L; --i) {
        while (j != R + 2 && s[j - 1] != s[i]) j = bor[j];
        bor[i] = --j;
        assert(j > i);
        if (j == R + 1)
          ju[i] = i, der[i] = 0;
        else
          der[i] = j - i, ju[i] = der[i] == der[j] ? ju[j] : i;
      }
    }
    for (int i = 1; i <= sn; ++i) ha[i] = (ha[i - 1] * B + s[i]) % mod;
    for (int i = 1; i <= m; ++i)
      SA(rp[i] - lp[i] + 1, s + lp[i] - 1, rk + lp[i] - 1, sa + lp[i] - 1);
    for (int i = 1; i <= m; ++i)
      for (int j = lp[i]; j <= rp[i]; ++j) sa[j] += lp[i] - 1, rk[sa[j]] = j;
    dfs(1);
  }
  inline int llen(int L, int R, int k) {
    int l = 0, r = min(R - L + 1, sb[k] - k + 1), mid;
    while (l != r) {
      mid = (l + r + 1) >> 1;
      if (ha1(k, k + mid - 1) == ha2(L, L + mid - 1))
        l = mid;
      else
        r = mid - 1;
    }
    return l;
  }
  inline bool cmp(int L, int R, int k) {
    k = sa[k];
    int x = llen(L, R, k);
    if (x == R - L + 1 || x == sb[k] - k + 1) return 1;
    return s[k + x] < s1[L + x];
  }
  inline int gpo(int L, int R, int k) {
    int l = lp[k] - 1, r = rp[k], mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (cmp(L, R, mid))
        l = mid;
      else
        r = mid - 1;
    }
    if (l == rp[k]) return sa[l];
    if (l == lp[k] - 1) return sa[l + 1];
    int x = sa[l], y = sa[l + 1];
    return llen(L, R, x) < llen(L, R, y) ? y : x;
  }
  inline int gd(int x, int R, int lm) {
    if (lm == 0) return 0;
    while (x <= R && R - ju[x] + 1 > lm) x = bor[ju[x]];
    if (x > R) return 0;
    int dd = der[x];
    int ln = R - x + 1;
    if (ln <= lm) return x;
    int kk = (ln - lm - 1) / dd + 1;
    return x + kk * dd;
  }
  inline void dfs(int x, int f = 0, int d = 0) {
    for (auto &h : oh[x]) {
      int L = dep[h.a], xh = h.c, k = h.b;
      int p = gpo(L, d, k), len = llen(L, d, p);
      pp[xh][rv] = gd(p, rp[k], len);
      p = pp[xh][rv];
    }
    for (register int i = fir[x]; i; i = nxt[i])
      if (to[i] != f) {
        s1[d + 1] = co[i], hs[d + 1] = (hs[d] * B + co[i]) % mod;
        dfs(to[i], x, d + 1);
      }
  }
} B1, B2;
inline int gcd(int a, int b) {
  int c;
  for (; b; c = a % b, a = b, b = c)
    ;
  return a;
}
inline void exgcd(int a, int b, long long &x, long long &y) {
  if (!b) return x = a, y = 0, void();
  exgcd(b, a % b, x, y);
  long long tmp = x;
  x = y, y = tmp - a / b * y;
}
inline int fc(int x, int y, int z, int l1, int l2) {
  if (z < 0) return 0;
  if (x == 0 && y == 0) return z == 0;
  if (x == 0) return z % y == 0 && z / y <= l2;
  if (y == 0) return z % x == 0 && z / x <= l1;
  int X = gcd(x, y);
  if (z % X) return 0;
  x /= X, y /= X, z /= X;
  long long A, B;
  exgcd(x, y, A, B);
  A = (A * z % y + y) % y;
  B = (z - A * x) / y;
  if (B < 0) return 0;
  assert(A * x + B * y == z);
  if (l1 < A) return 0;
  int U = min((l1 - A) / y, B / x), D = B <= l2 ? 0 : (B - l2 - 1) / x + 1;
  if (U < D) return 0;
  return U - D + 1;
}
inline int ask(int x, int y, int h) {
  int *t1 = B1.ju, *t2 = B2.ju, *d1 = B1.der, *d2 = B2.der;
  int L1 = sb[x], L2 = sb[y], res = 0;
  for (int i = x; i <= L1; i = B1.bor[t1[i]])
    for (int j = y; j <= L2; j = B2.bor[t2[j]]) {
      res += fc(d1[i], d2[j], L1 - i + 1 + L2 - j + 1 - h,
                d1[i] ? (t1[i] - i) / d1[i] : -1,
                d2[j] ? (t2[j] - j) / d2[j] : -1);
    }
  return res;
}
int main() {
  bin[0] = 1;
  for (int i = 1; i < mxn; ++i) bin[i] = bin[i - 1] * B % mod;
  a14 = scanf("%d%d%d", &n, &m, &qn);
  {
    char hh[5];
    for (int i = 1, u, v; i < n; ++i)
      a14 = scanf("%d%d%s", &u, &v, hh), gadd(u, v, hh[0]), gadd(v, u, hh[0]);
  }
  for (int i = 1, j = 1; i <= m; ++i)
    a14 = scanf("%s", ss + j), lp[i] = j, j += strlen(ss + j), rp[i] = j - 1;
  for (int i = 1; i <= m; ++i)
    for (int j = lp[i]; j <= rp[i]; ++j) bs[j] = i, sb[j] = rp[i];
  for (int i = 1; i <= qn; ++i) a14 = scanf("%d%d%d", a1 + i, a2 + i, a3 + i);
  dfs1(1, 0), dfs2(1, 0);
  for (int t = 1; t <= qn; ++t) {
    int x = a1[t], y = a2[t], z = lca(x, y), k = a3[t], len = rp[k] - lp[k] + 1;
    a4[t] = z, a5[t] = len;
    if (dep[x] - dep[z] >= len) {
      int u = kzx(x, dep[x] - dep[z] - len + 1);
      A1.vc[u].push_back((thr){k, -1, t}), A1.vc[x].push_back((thr){k, 1, t});
    }
    if (dep[y] - dep[z] >= len) {
      int u = kzx(y, dep[y] - dep[z] - len + 1);
      A2.vc[u].push_back((thr){k, -1, t}), A2.vc[y].push_back((thr){k, 1, t});
    }
    if (x != z && y != z) {
      B1.oh[x].push_back((thr){z, k, t}), B2.oh[y].push_back((thr){z, k, t});
    }
  }
  A1.w1(), A2.w2();
  B1.W(1), B2.W(0);
  for (int t = 1; t <= qn; ++t)
    if (pp[t][0] && pp[t][1]) qa[t] += ask(pp[t][1], pp[t][0], a5[t]);
  for (int i = 1; i <= qn; ++i) printf("%d\n", qa[i]);
  return 0;
}