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 = 101000;
const int Mod = 998244353;
class Data {
 public:
  int l, r, id;
};
int n, Q, m, K;
vector<Data> Qk[maxN];
int E[maxN], Ans[maxN], Ful[maxN], Cnt[maxN], Dwn[maxN];
int Inv[maxN + maxN], Bl[maxN], mul = 0;
bool qcmp(Data A, Data B);
void Add(int p);
void Del(int p);
int main() {
  Inv[0] = Inv[1] = 1;
  for (int i = 2; i < maxN + maxN; i++)
    Inv[i] = (Mod - 1ll * (Mod / i) * Inv[Mod % i] % Mod + Mod) % Mod;
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &E[i]), ++Ful[E[i]];
  int mxk = 0, Block = 1010;
  for (int i = 1; i <= Q; i++) {
    int l, r, k;
    scanf("%d%d%d", &l, &r, &k);
    Qk[k].push_back((Data){l, r, i});
    mxk = max(mxk, k);
  }
  for (int i = 1; i <= n; i++) Bl[i] = i / Block + 1;
  int c = 0;
  for (K = 0; K <= mxk; K++)
    if (Qk[K].size()) {
      int l = 1, r = 0;
      mul = 1;
      Dwn[n] = 1;
      for (int j = n - 1; j >= 1; j--)
        Dwn[j] = 1ll * Dwn[j + 1] * (1ll * m * K % Mod + n - j) % Mod;
      sort(Qk[K].begin(), Qk[K].end(), qcmp);
      int d = 0;
      for (int j = 0, sz = Qk[K].size(); j < sz; j++) {
        int ql = Qk[K][j].l, qr = Qk[K][j].r;
        d += abs(l - ql);
        d += abs(r - qr);
        while (r < qr) Add(++r);
        while (l > ql) Add(--l);
        while (r > qr) Del(r--);
        while (l < ql) Del(l++);
        Ans[Qk[K][j].id] = 1ll * mul * Dwn[qr - ql + 1] % Mod;
      }
      for (int j = l; j <= r; j++) --Cnt[E[j]];
    }
  for (int i = 1; i <= Q; i++) printf("%d\n", Ans[i]);
  return 0;
}
bool qcmp(Data A, Data B) {
  if (Bl[A.l] == Bl[B.l]) return A.r < B.r;
  return A.l < B.l;
}
void Add(int p) {
  mul = 1ll * mul * (Ful[E[p]] + K - Cnt[E[p]]) % Mod;
  ++Cnt[E[p]];
  return;
}
void Del(int p) {
  --Cnt[E[p]];
  mul = 1ll * mul * Inv[Ful[E[p]] + K - Cnt[E[p]]] % Mod;
  return;
}