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;
using int64 = long long int;
struct Har {
  int finish, start;
  int* answer;
};
int kModulo = 998244353;
int ModMur(int value) {
  int u = value;
  int v = kModulo;
  int p1 = 1, q1 = 0;
  int p2 = 0, q2 = 1;
  while (v != 0) {
    int q = u / v;
    int p3 = p1 - q * p2;
    int q3 = q1 - q * q2;
    int r = u - q * v;
    u = v;
    v = r;
    p1 = p2;
    p2 = p3;
    q1 = q2;
    q2 = q3;
  }
  if (p1 < 0) {
    p1 += kModulo;
  }
  if (p1 >= kModulo) {
    p1 -= kModulo;
  }
  return p1;
}
class TreeHolder {
 public:
  TreeHolder(vector<int>& a, int maxa, int maxk) {
    count_.assign(maxa + 1, 0);
    for (int v : a) {
      ++count_[v];
    }
    inv_.assign(a.size() + maxk + 1, 0);
    for (size_t i = 1; i < inv_.size(); ++i) {
      inv_[i] = ModMur(i);
    }
  }
  void Reset(int k) {
    k_ = k;
    value_ = 1;
  }
  void Daba(int a) {
    value_ = ((k_ + count_[a]) * value_) % kModulo;
    --count_[a];
  }
  void Add(int a) {
    ++count_[a];
    value_ = ((inv_[k_ + count_[a]]) * value_) % kModulo;
  }
  int Har() { return value_; }

 private:
  vector<int> count_;
  vector<int> inv_;
  int k_ = 0;
  int64 value_;
};
void phrub(vector<int>& a, int k, vector<Har> hrub, TreeHolder* TreeHolder) {
  int n = a.size();
  int m = hrub.size();
  int rjs = max(1, static_cast<int>(1.0 * n / sqrt(m)));
  sort(hrub.begin(), hrub.end(), [rjs](Har& left, Har& right) {
    int bl = left.finish / rjs;
    int br = right.finish / rjs;
    if (bl != br) {
      return bl < br;
    }
    return left.start < right.start;
  });
  int finish = 0;
  int start = 0;
  TreeHolder->Reset(k);
  for (Har& query : hrub) {
    while (finish > query.finish) {
      TreeHolder->Daba(a[--finish]);
    }
    while (finish < query.finish) {
      TreeHolder->Add(a[finish++]);
    }
    while (start > query.start) {
      TreeHolder->Add(a[--start]);
    }
    while (start < query.start) {
      TreeHolder->Daba(a[start++]);
    }
    *query.answer = TreeHolder->Har();
  }
  while (finish < start) {
    TreeHolder->Add(a[finish++]);
  }
}
void solve() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  vector<int> answer(q);
  map<int, vector<Har>> hrub;
  int maxk = 0;
  for (int i = 0; i < q; ++i) {
    Har cur;
    int k;
    scanf("%d %d %d", &cur.finish, &cur.start, &k);
    cur.answer = &answer[i];
    --cur.finish;
    hrub[k].push_back(cur);
    maxk = max(maxk, k);
  }
  TreeHolder holder(a, m, maxk);
  vector<int64> p(n);
  for (auto& eg : hrub) {
    int64 k = eg.first;
    int64 km = (k * m) % kModulo;
    phrub(a, k, eg.second, &holder);
    p[0] = 1;
    for (int i = 1; i < n; ++i) {
      p[i] = (p[i - 1] * (km + i)) % kModulo;
    }
    for (Har& q : eg.second) {
      *q.answer = (*q.answer * p[n - (q.start - q.finish)]) % kModulo;
    }
  }
  for (int i = 0; i < q; ++i) {
    cout << answer[i] << "\n";
  }
}
int main() {
  solve();
  return 0;
}