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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const int N = 100005, BAO = 200005, mod = 998244353, B = 200;
int n, m, Q, ans[N], e[N], q[N], tong[N], tt[N];
vector<int> v[N];
map<int, int> M[N];
long long fac[N << 1], ni[N << 1];
void ins(int id) {
  if (--M[tong[e[id]]][tt[e[id]]] == 0) M[tong[e[id]]].erase(tt[e[id]]);
  M[tong[e[id]]][++tt[e[id]]]++;
}
void del(int id) {
  if (--M[tong[e[id]]][tt[e[id]]] == 0) M[tong[e[id]]].erase(tt[e[id]]);
  M[tong[e[id]]][--tt[e[id]]]++;
}
void solve(int id) {
  v[id].resize(n);
  v[id][0] = 1;
  long long t = (long long)m * id % mod;
  for (int i = 1; i < n; i++) v[id][i] = v[id][i - 1] * (t + i) % mod;
}
struct data {
  int l, r, k, id;
} a[N];
bool operator<(data a, data b) {
  return a.l / B == b.l / B ? (((a.l / B) & 1) ? (a.r > b.r) : (a.r < b.r))
                            : a.l < b.l;
}
inline int ksm(long long a, int b) {
  int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans;
}
int p(int x, int y) { return fac[x] * ni[x - y] % mod; }
int main() {
  n = read();
  m = read();
  Q = read();
  for (int i = 1; i <= n; i++) tong[e[i] = read()]++;
  for (int i = 1; i <= m; i++) q[i] = tong[i];
  sort(&q[1], &q[m + 1]);
  int tot = unique(&q[1], &q[m + 1]) - q - 1;
  int l = 1, r = 0;
  for (int i = 1; i <= Q; i++) {
    a[i].l = read();
    a[i].r = read();
    a[i].k = read();
    a[i].id = i;
    if (!v[a[i].k].size()) solve(a[i].k);
    ans[i] = v[a[i].k][n - (a[i].r - a[i].l + 1)];
  }
  sort(&a[1], &a[Q + 1]);
  for (int i = fac[0] = 1; i <= BAO; i++) fac[i] = fac[i - 1] * i % mod;
  ni[BAO] = ksm(fac[BAO], mod - 2);
  for (int i = BAO; i; i--) ni[i - 1] = ni[i] * i % mod;
  for (int i = 1; i <= m; i++) M[tong[i]][0]++;
  for (int i = 1; i <= Q; i++) {
    while (l > a[i].l) ins(--l);
    while (r < a[i].r) ins(++r);
    while (l < a[i].l) del(l++);
    while (r > a[i].r) del(r--);
    long long dq = 1;
    for (int j = 1; j <= tot; j++) {
      for (auto k : M[q[j]]) {
        dq = dq * ksm(p(q[j] + a[i].k, k.first), k.second) % mod;
      }
    }
    ans[a[i].id] = ans[a[i].id] * dq % mod;
  }
  for (int i = 1; i <= Q; i++) writeln(ans[i]);
}