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 long long mod = 998244353;
const int N = 2e5 + 25;
const int M = 3e2 + 32;
int e[N], freq[N];
long long inv[N], npr[N], ans[N];
vector<array<int, 3> > query[N];
long long inverse(long long n) {
  if (n == 1) return 1;
  if (n >= mod) return inverse(n % mod);
  return (mod - mod / n) * inverse(mod % n) % mod;
}
void pre() {
  for (int i = 1; i < N; i++) {
    inv[i] = inverse(i);
  }
}
inline bool cmp(array<int, 3> &a, array<int, 3> &b) {
  int blkl_a = a[0] / M;
  int blkl_b = b[0] / M;
  if (blkl_a != blkl_b) return blkl_a < blkl_b;
  return a[1] < b[1];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(32);
  pre();
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    cin >> e[i];
  }
  for (int i = 1; i <= q; i++) {
    int l, r, k;
    cin >> l >> r >> k;
    query[k].push_back({l, r, i});
  }
  for (int i = 0; i < N; i++) {
    if (query[i].empty()) continue;
    sort(query[i].begin(), query[i].end(), cmp);
    long long cur_prod = 1;
    for (int j = 1; j <= m; j++) {
      freq[j] = i;
    }
    long long tot = i * 1LL * m % mod;
    npr[n] = 1;
    for (int j = 1; j <= n; j++) {
      freq[e[j]]++;
      npr[n - j] = npr[n - j + 1] * (tot + j) % mod;
    }
    int lo = 1, hi = 1;
    for (auto qq : query[i]) {
      int l = qq[0];
      int r = qq[1];
      int ind = qq[2];
      while (lo > l) {
        lo--;
        (cur_prod *= freq[e[lo]]) %= mod;
        freq[e[lo]]--;
      }
      while (hi <= r) {
        (cur_prod *= freq[e[hi]]) %= mod;
        freq[e[hi]]--;
        hi++;
      }
      while (lo < l) {
        freq[e[lo]]++;
        (cur_prod *= inv[freq[e[lo]]]) %= mod;
        lo++;
      }
      while (hi > r + 1) {
        hi--;
        freq[e[hi]]++;
        (cur_prod *= inv[freq[e[hi]]]) %= mod;
      }
      ans[ind] = cur_prod * npr[r - l + 1] % mod;
    }
  }
  for (int i = 1; i <= q; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}