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 MOD = 998244353;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 2e5 + 5;
int iv[maxn];
int fac[maxn];
int ifac[maxn];
int n, m, q;
int a[maxn];
vector<pair<pair<int, int>, int> > que[maxn];
int f[maxn];
int g[maxn];
int res[maxn];
int num;
int den;
int cnt[maxn];
inline void add(int x) { num = (long long)num * (cnt[x]--) % MOD; }
inline void rem(int x) { den = (long long)den * (++cnt[x]) % MOD; }
void work(int k) {
  if (!int((que[k]).size())) return;
  for (int i = (0); i < (m); ++i) g[i] = f[i] + k;
  long long s = 0;
  for (int i = (0); i < (m); ++i) s += g[i];
  static int x[maxn];
  x[0] = 1;
  for (int i = (0); i < (maxn - 1); ++i) x[i + 1] = mult(x[i], (s - i) % MOD);
  int magic = n / sqrt(int((que[k]).size())) + 1;
  sort((que[k]).begin(), (que[k]).end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         if (a.first.first / magic != b.first.first / magic)
           return a.first.first / magic < b.first.first / magic;
         return a.first.second < b.first.second;
       });
  num = den = 1;
  fill_n(cnt, maxn, 0);
  for (int i = (0); i < (m); ++i) cnt[i] = g[i];
  int l = 0, r = -1;
  for (auto e : que[k]) {
    int u = e.first.first;
    int v = e.first.second;
    int ix = e.second;
    int d = v - u + 1;
    while (l > u) add(a[--l]);
    while (r < v) add(a[++r]);
    while (l < u) rem(a[l++]);
    while (r > v) rem(a[r--]);
    res[ix] = mult(num, inv(den));
    res[ix] = mult(res[ix], x[n]);
    res[ix] = mult(res[ix], inv(x[d]));
  }
}
void chemthan() {
  for (int i = (1); i < (maxn); ++i) iv[i] = inv(i);
  fac[0] = 1;
  for (int i = (1); i < (maxn); ++i) fac[i] = mult(fac[i - 1], i);
  for (int i = (0); i < (maxn); ++i) ifac[i] = inv(fac[i]);
  cin >> n >> m >> q;
  for (int i = (0); i < (n); ++i) cin >> a[i], a[i]--, f[a[i]]++;
  for (int i = (0); i < (q); ++i) {
    int l, r, k;
    cin >> l >> r >> k;
    l--, r--;
    que[k].push_back(make_pair(make_pair(l, r), i));
  }
  for (int k = (0); k < (maxn); ++k) work(k);
  for (int i = (0); i < (q); ++i) cout << res[i] << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}