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 MAXN = 200005;
const int P = 998244353;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
struct querys {
  int l, r, k, home;
} qry[MAXN];
int prod[MAXN], inv[MAXN];
int n, m, q, size, blocks, block[MAXN];
int now, nowk, a[MAXN], b[MAXN], e[MAXN], ans[MAXN];
bool cmp(querys a, querys b) { return a.k < b.k; }
bool cnp(querys a, querys b) {
  if (block[a.l] == block[b.l])
    return a.r < b.r;
  else
    return block[a.l] < block[b.l];
}
int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0)
    return 1ll * tmp * tmp % P;
  else
    return 1ll * tmp * tmp % P * x % P;
}
void add(int col) {
  b[col]++;
  int tmp = a[col] + nowk - b[col] + 1;
  now = 1ll * now * tmp % P;
}
void del(int col) {
  int tmp = a[col] + nowk - b[col] + 1;
  now = 1ll * now * inv[tmp] % P;
  b[col]--;
}
int main() {
  read(n), read(m), read(q);
  for (int i = 1; i <= n; i++) read(e[i]), a[e[i]]++;
  for (int i = 1; i <= q; i++)
    read(qry[i].l), read(qry[i].r), read(qry[i].k), qry[i].home = i;
  for (int i = 1; i < MAXN; i++) inv[i] = power(i, P - 2);
  sort(qry + 1, qry + q + 1, cmp);
  for (int i = 1, nxt; i <= q; i = nxt) {
    nxt = i;
    while (nxt <= q && qry[nxt].k == qry[i].k) nxt++;
    int cntq = nxt - i, tmp = 1ll * qry[i].k * m % P;
    size = max(n / sqrt(cntq), 1.0), blocks = 0;
    prod[0] = 1;
    for (int j = 1; j <= n; j++) {
      prod[j] = 1ll * prod[j - 1] * (tmp + j) % P;
      if (j % size == 1 % size) blocks++;
      block[j] = blocks;
    }
    sort(qry + i, qry + nxt, cnp);
    int l = qry[i].l, r = qry[i].l - 1;
    now = 1, nowk = qry[i].k;
    memset(b, 0, sizeof(b));
    for (int j = i; j < nxt; j++) {
      while (r < qry[j].r) add(e[++r]);
      while (l > qry[j].l) add(e[--l]);
      while (r > qry[j].r) del(e[r--]);
      while (l < qry[j].l) del(e[l++]);
      ans[qry[j].home] = 1ll * now * prod[n - (qry[j].r - qry[j].l + 1)] % P;
    }
  }
  for (int i = 1; i <= q; i++) writeln(ans[i]);
  return 0;
}