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 N = 100010, Mo = 998244353;
int k, sum;
int w[N], c[N], rev[N * 2], a[N], pos[N], ans[N], f[N];
inline int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch == '-' ? o = -1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return o * x;
}
struct Dat {
  int l, r, t;
  inline bool operator<(const Dat &A) const {
    return pos[l] == pos[A.l] ? r < A.r : l < A.l;
  }
} g[N];
vector<Dat> q[N];
inline void add(int x) { sum = 1LL * sum * (w[x] + k - c[x]) % Mo, ++c[x]; }
inline void dec(int x) { --c[x], sum = 1LL * sum * rev[w[x] + k - c[x]] % Mo; }
inline void solve(vector<Dat> &g) {
  sort(g.begin(), g.end());
  int l = 1, r = 0;
  sum = 1;
  for (auto x : g) {
    while (l > x.l) add(a[--l]);
    while (r < x.r) add(a[++r]);
    while (l < x.l) dec(a[l++]);
    while (r > x.r) dec(a[r--]);
    ans[x.t] = 1LL * sum * f[r - l + 1] % Mo;
  }
}
int main() {
  int n, m, Q, s = 3000;
  cin >> n >> m >> Q;
  for (int i = 1; i <= n; i++) ++w[a[i] = gi()], pos[i] = (i - 1) / s;
  for (int i = 1; i <= Q; i++) {
    int x = gi(), y = gi(), z = gi();
    q[z].push_back((Dat){x, y, i});
  }
  rev[1] = 1;
  for (int i = 2; i <= n + 100000; i++)
    rev[i] = -1LL * Mo / i * rev[Mo % i] % Mo;
  for (k = 0; k <= 100000; k++)
    if (q[k].size()) {
      int s = 1LL * k * m % Mo;
      f[n] = 1;
      memset(c, 0, sizeof(c));
      for (int i = n; i; i--) f[i - 1] = 1LL * f[i] * (s + n - i + 1) % Mo;
      solve(q[k]);
    }
  for (int i = 1; i <= Q; i++) printf("%d\n", (ans[i] + Mo) % Mo);
  return 0;
}