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 mo = 998244353, N = 1e5 + 5;
int n, m, q, e[N], l, r, k, ans[N], i, j, a[N], b[N], ssm[N];
vector<vector<int> > qu[N];
int bel[N], bl, inv[N * 2], pi, xjm[N];
inline void inc(int i) {
  i = e[i];
  pi = 1ll * pi * (a[i] + k - b[i]++) % mo;
}
inline void dec(int i) {
  i = e[i];
  pi = 1ll * pi * inv[a[i] + k - (--b[i])] % mo;
}
int main() {
  for (i = 2, inv[1] = 1; i < N * 2; ++i)
    inv[i] = 1ll * (mo - mo / i) * inv[mo % i] % mo;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (i = 1; i <= n; ++i) cin >> e[i], ++a[e[i]];
  for (i = 1; i <= q; ++i) cin >> l >> r >> k, qu[k].push_back({l, r, i});
  for (k = 0; k < N; ++k)
    if (!qu[k].empty()) {
      bl = max(n / sqrt(qu[k].size()), 1.0);
      for (i = 1; i <= n; ++i) bel[i] = (i - 1) / bl + 1;
      sort(qu[k].begin(), qu[k].end(),
           [&](const vector<int>& a, const vector<int>& b) {
             return bel[a[0]] == bel[b[0]]
                        ? (bel[a[0]] & 1 ? a[1] > b[1] : a[1] < b[1])
                        : a[0] < b[0];
           });
      j = 1ll * k * m % mo;
      ssm[0] = 1;
      for (i = 1; i <= n; ++i) ssm[i] = 1ll * ssm[i - 1] * (j + i) % mo;
      l = 1;
      r = 0;
      pi = 1;
      memset(b + 1, 0, m << 2);
      for (vector<int> u : qu[k]) {
        for (; l > u[0]; inc(--l))
          ;
        for (; r < u[1]; inc(++r))
          ;
        for (; l < u[0]; dec(l++))
          ;
        for (; r > u[1]; dec(r--))
          ;
        ans[u[2]] = 1ll * pi * ssm[n - (u[1] - u[0] + 1)] % mo;
      }
    }
  for (i = 1; i <= q; ++i) cout << ans[i] << '\n';
  return 0;
}