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 A[100005];
vector<pair<pair<int, int>, int>> que[100005];
const int DIV = 3162;
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  int aa = a.first.first / DIV, bb = b.first.first / DIV;
  if (aa != bb) return aa < bb;
  return a.first.second < b.first.second;
}
long long power, ans[100005];
int freq[100005];
const long long MOD = 998244353;
const int MAXN = 200005;
long long fac[MAXN], invfac[MAXN];
long long inv[MAXN];
void add(int p) {
  int v = A[p];
  power = power * freq[v] % MOD;
  --freq[v];
}
void rem(int p) {
  int v = A[p];
  ++freq[v];
  power = power * inv[freq[v]] % MOD;
}
int tot[MAXN];
long long tt[MAXN];
long long modex(long long a, long long p, long long m = MOD) {
  if (!p) return 1;
  if (p & 1) return a * modex(a, p - 1, m) % m;
  long long v = modex(a, p >> 1, m);
  return v * v % m;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i < MAXN; ++i) fac[i] = (fac[i - 1] * i) % MOD;
  invfac[MAXN - 1] = modex(fac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; --i)
    invfac[i] = (invfac[i + 1] * (i + 1)) % MOD;
  for (int i = 1; i < MAXN; ++i) inv[i] = invfac[i] * fac[i - 1] % MOD;
  int N, M, Q;
  scanf("%d%d%d", &N, &M, &Q);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    ++tot[A[i]];
  }
  for (int q = 0; q < Q; ++q) {
    int l, r, k;
    scanf("%d%d%d", &l, &r, &k);
    que[k].push_back({{l, r}, q});
  }
  for (int i = 0; i <= 100000; ++i) {
    auto &queries = que[i];
    if (queries.empty()) continue;
    sort(queries.begin(), queries.end(), cmp);
    tt[0] = 1;
    long long val = (long long)M * i % MOD;
    for (int j = 1; j <= 100000; ++j) {
      val = (val + 1) % MOD;
      freq[j] = i + tot[j];
      tt[j] = tt[j - 1] * val % MOD;
    }
    power = 1;
    int lef = 1, rig = 0;
    for (int q = 0; q < queries.size(); ++q) {
      int l = queries[q].first.first, r = queries[q].first.second;
      while (rig < r) add(++rig);
      while (rig > r) rem(rig--);
      while (lef < l) rem(lef++);
      while (lef > l) add(--lef);
      ans[queries[q].second] = power * tt[N - (r - l + 1)] % MOD;
    }
  }
  for (int q = 0; q < Q; ++q) printf("%lld\n", ans[q]);
}