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 = 2e5 + 50, mo = 998244353;
inline int gi() {
  char ch = getchar();
  int q = 0, x = 0;
  while (ch < '0' || ch > '9') q = (ch == '-' ? 1 : q), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int bl[N];
struct data {
  int l, r, x;
  bool operator<(const data &c) const {
    return bl[l] == bl[c.l] ? r < c.r : l < c.l;
  }
} w[N];
vector<data> q[N];
int a[N], f[N], c[N], g[N], b[N], rev[N], tot, ans;
void add(int x) {
  ans = 1ll * ans * (g[x] - c[x]) % mo, ++c[x];
  return;
}
void del(int x) {
  --c[x], ans = 1ll * ans * rev[g[x] - c[x]] % mo;
  return;
}
void solve() {
  sort(w + 1, w + tot + 1);
  int l = 1, r = 0;
  ans = 1;
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= tot; ++i) {
    while (l > w[i].l) add(a[--l]);
    while (r < w[i].r) add(a[++r]);
    while (l < w[i].l) del(a[l++]);
    while (r > w[i].r) del(a[r--]);
    b[w[i].x] = 1ll * ans * f[r - l + 1] % mo;
  }
  return;
}
int main() {
  int n = gi(), m = gi(), Q = gi(), x, y, z;
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i) ++g[a[i]];
  for (int i = 1; i <= Q; ++i)
    x = gi(), y = gi(), z = gi(), q[z].push_back((data){x, y, i});
  int blo = 3000;
  for (int i = 1; i <= n; ++i) bl[i] = i / blo;
  rev[0] = rev[1] = 1;
  for (int i = 2; i <= n + 100000; ++i)
    rev[i] = mo - 1ll * mo / i * rev[mo % i] % mo;
  for (int i = 0; i <= 100000; ++i)
    if (q[i].size()) {
      tot = 0;
      int sum = (1ll * m * i + n) % mo;
      for (int j = 0; j < q[i].size(); ++j) w[++tot] = q[i][j];
      f[n] = 1;
      for (int j = n; j; --j) f[j - 1] = 1ll * f[j] * (sum - j + 1) % mo;
      for (int j = 1; j <= m; ++j) g[j] += i;
      solve();
      for (int j = 1; j <= m; ++j) g[j] -= i;
    }
  for (int i = 1; i <= Q; ++i) printf("%d\n", b[i]);
  return 0;
}