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 P = 998244353, N = 100005;
using ll = long long;
int n, m, qq, ee[N], l, r, kk, ans[N], a[N], b[N], sum[N];
vector<vector<int>> q[N];
int bel[N], bl, tt[N];
ll pi, inv[N * 2];
inline void ins(int x) {
  x = ee[x];
  pi = pi * (a[x] + kk - b[x]++) % P;
}
inline void del(int x) {
  x = ee[x];
  pi = pi * inv[a[x] + kk - --b[x]] % P;
}
int main() {
  ios::sync_with_stdio(false);
  inv[1] = 1;
  for (int i = 2; i < N * 2; i++) inv[i] = (P - P / i) * inv[P % i] % P;
  cin >> n >> m >> qq;
  for (int i = 1; i <= n; i++) cin >> ee[i], ++a[ee[i]];
  for (int i = 1; i <= qq; i++) cin >> l >> r >> kk, q[kk].push_back({l, r, i});
  for (kk = 0; kk < N; kk++) {
    if (q[kk].empty()) continue;
    bl = max(n / sqrt(q[kk].size()), 1.0);
    for (int i = 1; i <= n; i++) bel[i] = (i - 1) / bl + 1;
    sort(q[kk].begin(), q[kk].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];
         });
    int j = 1ll * kk * m % P;
    sum[0] = 1;
    for (int i = 1; i <= n; i++) sum[i] = ll(sum[i - 1]) * (j + i) % P;
    l = 1;
    r = 0;
    pi = 1;
    memset(b + 1, 0, sizeof(int) * m);
    for (auto &i : q[kk]) {
      while (l > i[0]) ins(--l);
      while (r < i[1]) ins(++r);
      while (l < i[0]) del(l++);
      while (r > i[1]) del(r--);
      ans[i[2]] = pi * sum[n - (i[1] - i[0] + 1)] % P;
    }
  }
  for (int i = 1; i <= qq; i++) cout << ans[i] << endl;
  return 0;
}