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 MOD = 998244353;
const int N = 2e5 + 5;
const int MAXK = 1e5;
int n, m, q, nn;
vector<int> lst[N];
int ai[N], li[N], ri[N], ki[N], res[N];
int cnt[N], sum[N], inv[N], fwd[N], ans;
int comp(int a, int b) {
  if (li[a] / nn == li[b] / nn)
    return ri[a] < ri[b];
  else
    return li[a] / nn < li[b] / nn;
}
void add(int t) { ans = 1ll * ans * (cnt[t]--) % MOD; }
void rem(int t) { ans = 1ll * ans * inv[++cnt[t]] % MOD; }
int main() {
  scanf("%d%d%d", &n, &m, &q);
  nn = max(1.0, sqrt(1.0 * (n * 100) * (n * 100) / q));
  for (int i = 1; i <= n; ++i) scanf("%d", &ai[i]), sum[ai[i]]++;
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d%d", &li[i], &ri[i], &ki[i]);
    lst[ki[i]].push_back(i);
  }
  inv[1] = 1;
  for (int i = 2; i <= MAXK * 2; ++i)
    inv[i] = 1ll * (MOD - (MOD / i)) * inv[MOD % i] % MOD;
  for (int k = 0; k <= MAXK; ++k)
    if (lst[k].size()) {
      fwd[0] = 1;
      for (int i = 1; i <= n; ++i)
        fwd[i] = 1ll * fwd[i - 1] * (1ll * m * k % MOD + i) % MOD;
      sort(lst[k].begin(), lst[k].end(), comp);
      for (int i = 1; i <= m; ++i) cnt[i] = sum[i] + k;
      ans = 1;
      int L = 1, R = 0;
      for (int i : lst[k]) {
        while (R < ri[i]) add(ai[++R]);
        while (L > li[i]) add(ai[--L]);
        while (R > ri[i]) rem(ai[R--]);
        while (L < li[i]) rem(ai[L++]);
        res[i] = 1ll * fwd[n - (ri[i] - li[i] + 1)] * ans % MOD;
      }
    }
  for (int i = 1; i <= q; ++i) printf("%d\n", res[i]);
}