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 std::max;
using std::min;
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3f;
std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
inline _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
inline void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
inline void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
inline void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  f && (x = -x);
}
template <typename _Tp, typename... Args>
inline void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
inline int read_str(char *s) {
  char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  char *tar = s;
  *tar++ = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *tar++ = ch, ch = getchar();
  return *tar = 0, tar - s;
}
const int N = 200005;
const int mod = 998244353;
int e[N], a[N], inv[N];
struct info {
  int l, r, k, block, id;
  inline bool operator<(const info &o) const {
    return block == o.block ? block & 1 ? r > o.r : r < o.r : block < o.block;
  }
} q[N];
int n, m, b[N], ans, k, res[N], _[N];
inline void add(int x) { ans = 1LL * ans * (a[x] + k - (b[x]++)) % mod; }
inline void del(int x) { ans = 1LL * ans * inv[a[x] + k - (--b[x])] % mod; }
void solve(int L, int R) {
  ans = 1, k = q[L].k;
  int mul = 1;
  memset(b, 0, sizeof(b));
  for (int i = 0; i <= n; ++i)
    _[i] = mul, mul = 1LL * mul * ((1LL * k * m + i + 1) % mod) % mod;
  int B = max(1, (int)(n / sqrt(R - L + 1)));
  for (int i = L; i <= R; ++i) q[i].block = q[i].l / B;
  std::sort(q + L, q + R + 1);
  int l = 1, r = 0;
  for (int i = L; i <= R; ++i) {
    while (r < q[i].r) add(e[++r]);
    while (l > q[i].l) add(e[--l]);
    while (r > q[i].r) del(e[r--]);
    while (l < q[i].l) del(e[l++]);
    res[q[i].id] = 1LL * ans * _[n - (q[i].r - q[i].l + 1)] % mod;
  }
}
int main() {
  inv[0] = inv[1] = 1;
  for (int i = 2; i < N; ++i)
    inv[i] = 1LL * inv[mod % i] * (mod - mod / i) % mod;
  int Q;
  read(n, m, Q);
  for (int i = 1; i <= n; ++i) read(e[i]), ++a[e[i]];
  for (int i = 1; i <= Q; ++i) read(q[i].l, q[i].r, q[i].k), q[i].id = i;
  std::sort(q + 1, q + Q + 1,
            [&](const info &a, const info &b) -> bool { return a.k < b.k; });
  for (int i = 1; i <= Q; ++i) {
    int j = i;
    while (j < Q && q[j + 1].k == q[i].k) ++j;
    solve(i, j), i = j;
  }
  for (int i = 1; i <= Q; ++i) printf("%d\n", res[i]);
  return 0;
}