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;
int e[100000], c[100000], d[100000], other[100001];
int inv(long long int n) {
  long long int r = 1;
  int e = 998244353 - 2;
  while (e > 0) {
    if (e & 1) r *= n, r %= 998244353;
    e >>= 1;
    n *= n, n %= 998244353;
  }
  return r;
}
int invn[200001];
struct query {
  int l, r, i;
};
vector<query> v[100001];
int B;
bool comp(query a, query b) {
  if (a.l / B == b.l / B) {
    if ((a.l / B) & 1)
      return a.r > b.r;
    else
      return a.r < b.r;
  } else
    return a.l / B < b.l / B;
}
int ans[100000];
long long int aa;
int add(int p, int k) {
  aa *= c[e[p]] + k - d[e[p]], aa %= 998244353, d[e[p]]++;
  return 0;
}
int del(int p, int k) {
  d[e[p]]--, aa *= invn[c[e[p]] + k - d[e[p]]], aa %= 998244353;
  return 0;
}
int main() {
  int i;
  int n, m, q;
  int l, r, k;
  scanf("%d %d %d", &n, &m, &q);
  for (i = 0; i < n; i++) scanf("%d", &e[i]), c[--e[i]]++;
  for (i = 0; i < q; i++)
    scanf("%d %d %d", &l, &r, &k), v[k].push_back((query){l - 1, r - 1, i});
  int j;
  for (i = 1; i <= 200000; i++) invn[i] = inv(i);
  for (i = 0; i <= 100000; i++) {
    if (v[i].empty()) continue;
    B = n / sqrt(v[i].size()) + 1;
    sort(v[i].begin(), v[i].end(), comp);
    int l = 0, r = -1;
    other[n] = 1;
    for (j = n - 1; j >= 1; j--)
      other[j] = (other[j + 1] * (((long long int)i * m + n - j) % 998244353)) %
                 998244353;
    aa = 1;
    for (j = 0; j < v[i].size(); j++) {
      while (l > v[i][j].l) add(--l, i);
      while (r < v[i][j].r) add(++r, i);
      while (l < v[i][j].l) del(l++, i);
      while (r > v[i][j].r) del(r--, i);
      ans[v[i][j].i] = (aa * other[r - l + 1]) % 998244353;
    }
    fill(d, d + m, 0);
  }
  for (i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}