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>
#pragma comment(linker, "/STACK:512000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = (1e18) + 7;
const long double eps = 1e-9;
const long long mod = 998244353LL;
const int sq = 1000;
struct pyt {
  int l, p, k, num;
  pyt(int L, int P, int K, int N) {
    l = L;
    p = P;
    k = K;
    num = N;
  }
  bool operator<(const pyt &a) const {
    if (l / sq != a.l / sq) return l < a.l;
    return p < a.p;
  }
};
vector<long long> licz[maxn];
int n, m, q;
int odp[maxn];
int ile[maxn], cnt[maxn];
int tab[maxn];
unordered_map<long long, int> mapa, xd;
unordered_set<long long> zb;
int total[maxn];
vector<pyt> V;
inline void dodaj(int x) {
  x = tab[x];
  if (xd[total[x] + (1LL << 32) * ile[x]]-- == 1)
    xd.erase(total[x] + (1LL << 32) * ile[x]);
  ile[x]++;
  xd[total[x] + (1LL << 32) * ile[x]]++;
}
inline void usun(int x) {
  x = tab[x];
  if (xd[total[x] + (1LL << 32) * ile[x]]-- == 1)
    xd.erase(total[x] + (1LL << 32) * ile[x]);
  ile[x]--;
  xd[total[x] + (1LL << 32) * ile[x]]++;
}
long long pot(long long a, long long w) {
  a %= mod;
  if (w == 0) return 1;
  long long b = pot(a, w / 2);
  b = (b * b) % mod;
  if (w & 1) b = (b * a) % mod;
  return b;
}
map<pair<long long, long long>, int> pamie;
long long liczba(long long a, long long b) {
  if (!pamie.count({a, b})) {
    long long res = 1LL;
    b = a - b;
    while (a > b) {
      res = (res * (a % mod)) % mod;
      a--;
    }
    pamie[{a, b}] = res;
  }
  return pamie[{a, b}];
}
long long sil[maxn], odw[maxn];
inline long long tipoci(long long t, long long c) {
  return (sil[t] * odw[t - c]) % mod;
}
const long long c = ((1LL << 32) - 1LL);
inline long long calc(int k) {
  long long res = 1LL;
  for (auto s : xd) {
    assert(s.second);
    res = (res * pot(tipoci((s.first & c) + (long long)k, s.first >> 32LL),
                     s.second)) %
          mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sil[0] = 1;
  odw[0] = 1;
  for (int i = 1; i < maxn; i++)
    sil[i] = (sil[i - 1] * (long long)i) % mod, odw[i] = pot(sil[i], mod - 2);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> tab[i], total[tab[i]]++;
  for (int i = 1; i <= m; i++) xd[total[i]]++, zb.insert(total[i]);
  for (int i = 1; i <= q; i++) {
    int l, p, k;
    cin >> l >> p >> k;
    V.push_back(pyt(l, p, k, i));
    licz[k].push_back(1);
  }
  for (int i = 0; i < maxn; i++)
    if (((int)(licz[i]).size()) >= 1) {
      licz[i].clear();
      licz[i].push_back(1);
      for (int dl = 1; dl <= n; dl++)
        licz[i].push_back((licz[i].back() * ((m * (long long)i + dl) % mod)) %
                          mod);
      reverse(licz[i].begin(), licz[i].end());
    }
  sort(V.begin(), V.end());
  int l = 1;
  int p = 0;
  for (auto s : V) {
    while (l > s.l) dodaj(--l);
    while (p < s.p) dodaj(++p);
    while (l < s.l) usun(l++);
    while (p > s.p) usun(p--);
    odp[s.num] = (calc(s.k) * (long long)(licz[s.k][p - l + 1])) % mod;
  }
  for (int i = 1; i <= q; i++) cout << odp[i] << "\n";
  return 0;
}