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;
using LL = long long;
const int mod = 998244353;
int inv(int x) { return x < 2 ? x : (LL)mod / x * (mod - inv(mod % x)) % mod; }
const int N = 1e5 + 10, M = 200;
int ch[N << 1], f[N], c[N], rev[N << 1];
int a[N], L[N], R[N], K[N], I[N], ans[N], t[N];
int n, m, q, val, cL, cR;
inline void add(const int& i) {
  if (t[i] > c[i]) val = (LL)val * (t[i] - c[i]) % mod;
  c[i]++;
}
inline void sub(const int& i) {
  val = (LL)val * rev[t[i] - c[i] + 1] % mod;
  c[i]--;
}
inline void change(const int& L, const int& R) {
  if (cR < L || cL > R) {
    for (; cL <= cR; sub(a[cL++]))
      ;
    for (cL = L, cR = L - 1; cR < R; add(a[++cR]))
      ;
  } else {
    for (; cL < L; sub(a[cL++]))
      ;
    for (; cL > L; add(a[--cL]))
      ;
    for (; cR < R; add(a[++cR]))
      ;
    for (; cR > R; sub(a[cR--]))
      ;
  }
}
void solve(int* id, int sz, int k) {
  int com = 1;
  for (int i = 0; i < m; i++) t[i] = k, c[i] = 0;
  for (int i = 0; i < n; i++) t[a[i]]++;
  for (int i = f[0] = 1; i <= n; i++)
    f[i] = f[i - 1] * ((LL)m * k % mod + i) % mod;
  for (int i = 0; i < m; i++) com = (LL)com * ch[t[i]] % mod;
  cL = 1, cR = 0, val = inv(com);
  for (int i = 0; i < sz; i++) {
    change(L[id[i]], R[id[i]]);
    ans[id[i]] = (LL)val * com % mod * f[n - (R[id[i]] - L[id[i]] + 1)] % mod;
  }
}
int main() {
  for (int i = ch[0] = 1; i < N << 1; i++)
    ch[i] = (LL)i * ch[i - 1] % mod, rev[i] = inv(i);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    a[i]--;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", L + i, R + i, K + i);
    L[i]--, R[i]--;
    I[i] = i;
  }
  sort(I, I + q, [&](int i, int j) {
    if (K[i] != K[j]) return K[i] < K[j];
    if (L[i] / M != L[j] / M) return L[i] < L[j];
    if ((L[i] / M) & 1) return R[i] > R[j];
    return R[i] < R[j];
  });
  for (int i = 0, j; i < q; i = j) {
    for (j = i; j < q && K[I[j]] == K[I[i]]; j++)
      ;
    solve(I + i, j - i, K[I[i]]);
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}