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 N = 2e5 + 10, M = 1e5 + 10, K = 1e5 + 10, Q = 1e5 + 10;
const long long MOD = 998244353;
struct Query {
  int id, l, r, k;
  void rd(int i) {
    id = i;
    scanf("%d%d%d", &l, &r, &k);
    --l;
  }
  bool operator<(const Query &rhs) const { return k < rhs.k; }
} q[Q];
long long fac[N], inv[N], ans[N];
int B, cnt[N], e[N], c[N], n, m, qn;
bool cmp(Query x, Query y) {
  int bx = x.l / B, by = y.l / B;
  if (bx != by) return bx < by;
  return bx & 1 ? x.r > x.r : x.r < y.r;
}
int main() {
  scanf("%d%d%d", &n, &m, &qn);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; ++i) {
    scanf("%d", &e[i]);
    ++cnt[e[i]];
  }
  for (int i = 0; i < qn; ++i) q[i].rd(i);
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= 200000; ++i) {
    inv[i] = -inv[MOD % i] * (MOD / i) % MOD;
    while (inv[i] < 0) inv[i] += MOD;
  }
  sort(q, q + qn);
  for (int ql = 0, qr = 0; ql < qn; ql = qr) {
    int k = q[ql].k, l = 0, r = 0;
    long long s = 1;
    for (qr = ql; qr < qn && q[qr].k == k; ++qr)
      ;
    fac[0] = 1;
    for (int i = 1; i <= n; ++i)
      fac[i] = (1ll * k * m + i) % MOD * fac[i - 1] % MOD;
    for (int i = 1; i <= m; ++i) c[i] = cnt[i] + k;
    B = n / sqrt(qr - ql) + 1;
    sort(q + ql, q + qr, cmp);
    for (int i = ql; i < qr; ++i) {
      while (l > q[i].l) s = s * c[e[--l]]-- % MOD;
      while (r < q[i].r) s = s * c[e[r++]]-- % MOD;
      while (l < q[i].l) s = s * inv[++c[e[l++]]] % MOD;
      while (r > q[i].r) s = s * inv[++c[e[--r]]] % MOD;
      ans[q[i].id] = s * fac[n - r + l] % MOD;
    }
  }
  for (int i = 0; i < qn; ++i) cout << ans[i] << endl;
  return 0;
}