Problem: In "The Man in the High Castle" world, there are m different film endings. 

Abendsen owns a storage and a shelf. At first, he has n ordered films on the shelf. In the i-th month he will do:

  1. Empty the storage.
  2. Put k_i ⋅ m films into the storage, k_i films for each ending.
  3. He will think about a question: if he puts all the films from the shelf into the storage, then randomly picks n films (from all the films in the storage) and rearranges them on the shelf, what is the probability that sequence of endings in [l_i, r_i] on the shelf will not be changed? Notice, he just thinks about this question, so the shelf will not actually be changed.



Answer all Abendsen's questions.

Let the probability be fraction P_i. Let's say that the total number of ways to take n films from the storage for i-th month is A_i, so P_i ⋅ A_i is always an integer. Print for each month P_i ⋅ A_i \pmod {998244353}.

998244353 is a prime number and it is equal to 119 ⋅ 2^{23} + 1.

It is guaranteed that there will be only no more than 100 different k values.

Input

The first line contains three integers n, m, and q (1 ≤ n, m, q ≤ 10^5, n+q≤ 10^5) — the number of films on the shelf initially, the number of endings, and the number of months.

The second line contains n integers e_1, e_2, …, e_n (1≤ e_i≤ m) — the ending of the i-th film on the shelf.

Each of the next q lines contains three integers l_i, r_i, and k_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ k_i ≤ 10^5) — the i-th query.

It is guaranteed that there will be only no more than 100 different k values.

Output

Print the answer for each question in a separate line.

Examples

Input

6 4 4
1 2 3 4 4 4
1 4 0
1 3 2
1 4 2
1 5 2


Output

6
26730
12150
4860


Input

5 5 3
1 2 3 4 5
1 2 100000
1 4 4
3 5 5


Output

494942218
13125
151632

Note

In the first sample in the second query, after adding 2 ⋅ m films into the storage, the storage will look like this: \{1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4\}.

There are 26730 total ways of choosing the films so that e_l, e_{l+1}, …, e_r will not be changed, for example, [1, 2, 3, 2, 2] and [1, 2, 3, 4, 3] are such ways.

There are 2162160 total ways of choosing the films, so you're asked to print (26730/2162160 ⋅ 2162160) mod 998244353 = 26730.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010, maxk = 100010, maxm = 100010, maxq = 100010,
          mod = 119 << 23 | 1;
int n, m, q, e[maxn], cnt[maxm], ans[maxq];
struct Query {
  int id, l, r, k;
  void rd(int i) {
    id = i;
    scanf("%d%d%d", &l, &r, &k);
    --l;
  }
  bool operator<(const Query& q) const { return k < q.k; }
} Q[maxq];
int B;
bool cmp(Query a, Query b) {
  int la = a.l / B, lb = b.l / B;
  if (la != lb) return la < lb;
  return la & 1 ? a.r > b.r : a.r < b.r;
}
int fac[maxn + maxk], iv[maxn + maxk];
long long inv(int a, int p = mod) {
  return a == 1 ? 1 : (1 + p * (a - inv(p % a, a))) / a % p;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n; i++) scanf("%d", e + i), --e[i];
  for (int i = 0; i < q; i++) Q[i].rd(i);
  std::sort(Q, Q + q);
  int lim = n + Q[q - 1].k;
  for (int i = *fac = 1; i <= lim; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  iv[lim] = inv(fac[lim]);
  for (int i = lim; i; i--) {
    iv[i - 1] = 1ll * iv[i] * i % mod;
    iv[i] = 1ll * fac[i - 1] * iv[i] % mod;
  }
  for (int ql = 0, qr = 0; ql < q; ql = qr) {
    int k = Q[ql].k;
    while (qr < q && Q[qr].k == k) qr++;
    for (int j = *fac = 1; j <= n; j++)
      fac[j] = (1ll * k * m + j) % mod * fac[j - 1] % mod;
    B = n / sqrt(qr - ql) + 1;
    std::sort(Q + ql, Q + qr, cmp);
    int l = Q[ql].l, r = Q[ql].l, s = 1;
    for (int i = 0; i < m; i++) cnt[i] = k;
    for (int i = 0; i < n; i++) cnt[e[i]]++;
    for (int i = ql; i < qr; i++) {
      while (l > Q[i].l) s = 1ll * s * cnt[e[--l]]-- % mod;
      while (r < Q[i].r) s = 1ll * s * cnt[e[r++]]-- % mod;
      while (l < Q[i].l) s = 1ll * s * iv[++cnt[e[l++]]] % mod;
      while (r > Q[i].r) s = 1ll * s * iv[++cnt[e[--r]]] % mod;
      ans[Q[i].id] = 1ll * s * fac[n - r + l] % mod;
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}