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 M = 998244353;
struct mint {
  int value;
  mint() : value(0) {}
  template <class T>
  mint(T x) {
    value = x % M;
    if (value < 0) {
      value += M;
    }
  }
  mint &operator++() {
    if (++value == M) {
      value = 0;
    }
    return *this;
  }
  mint &operator--() {
    if (--value == -1) {
      value = M - 1;
    }
    return *this;
  }
  mint operator++(int) {
    mint res = *this;
    ++*this;
    return res;
  }
  mint operator--(int) {
    mint res = *this;
    --*this;
    return res;
  }
  mint &operator+=(mint oth) {
    value += oth.value;
    if (value >= M) {
      value -= M;
    }
    return *this;
  }
  mint &operator-=(mint oth) {
    value -= oth.value;
    if (value < 0) {
      value += M;
    }
    return *this;
  }
  mint &operator*=(mint oth) {
    value = (long long)value * oth.value % M;
    return *this;
  }
  mint operator+() const { return *this; }
  mint operator-() const { return mint() - *this; }
  mint pow(long long e = M - 2) const {
    mint res = 1, b = *this;
    while (e > 0) {
      if (e % 2 == 1) {
        res *= b;
      }
      b *= b;
      e /= 2;
    }
    return res;
  }
  friend mint operator+(mint a, mint b) { return a += b; }
  friend mint operator-(mint a, mint b) { return a -= b; }
  friend mint operator*(mint a, mint b) { return mint(a) *= b; }
  friend bool operator==(mint a, mint b) { return a.value == b.value; }
  friend bool operator!=(mint a, mint b) { return a.value != b.value; }
  friend ostream &operator<<(ostream &out, mint a) {
    out << a.value;
    return out;
  }
};
const int N = 100000 + 1;
const int B = 512;
mint inv[2 * N], perm[N], ans[N], product;
int film[N], total[N], segment[N];
vector<array<int, 3>> queries[N];
void insert(int a) {
  a = film[a];
  product *= total[a] - segment[a];
  ++segment[a];
}
void erase(int a) {
  a = film[a];
  --segment[a];
  product *= inv[total[a] - segment[a]];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  inv[1] = 1;
  for (int i = 2; i < 2 * N; ++i) {
    inv[i] = -inv[M % i] * (M / i);
  }
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> film[i];
    ++total[film[i]];
  }
  for (int i = 1; i <= q; ++i) {
    int l, r, k;
    cin >> l >> r >> k;
    queries[k].push_back({l, r, i});
  }
  for (int i = 0; i < N; ++i) {
    if (queries[i].empty()) {
      continue;
    }
    sort(queries[i].begin(), queries[i].end(),
         [&](array<int, 3> a, array<int, 3> b) {
           if (a[0] / B == b[0] / B) {
             return a[1] < b[1];
           } else {
             return a[0] < b[0];
           }
         });
    for (int j = 1; j <= m; ++j) {
      total[j] += i;
      segment[j] = 0;
    }
    perm[0] = 1;
    for (int j = 1; j <= n; ++j) {
      perm[j] = perm[j - 1] * (mint(m) * i + j);
    }
    int l = 1, r = 0;
    product = 1;
    for (auto [a, b, id] : queries[i]) {
      while (l > a) {
        insert(--l);
      }
      while (r < b) {
        insert(++r);
      }
      while (l < a) {
        erase(l++);
      }
      while (r > b) {
        erase(r--);
      }
      ans[id] = product * perm[n - b + a - 1];
    }
    for (int j = 1; j <= m; ++j) {
      total[j] -= i;
    }
  }
  for (int i = 1; i <= q; ++i) {
    cout << ans[i] << "\n";
  }
}