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 GCC optimize("Ofast")
using namespace std;
const int maxn = 100010, P = 998244353;
int n, m, q, a[maxn], cnt[maxn], ans[maxn], occ[maxn], mul[maxn], tmp[200010];
struct query {
  int l, r, id;
};
vector<query> Q[maxn];
vector<int> val[maxn], inv[maxn];
int qp(int x, int y) {
  int z = 1;
  for (; y; y >>= 1, x = 1LL * x * x % P) {
    if (y & 1) z = 1LL * z * x % P;
  }
  return z;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]), cnt[a[i]]++;
  }
  for (int i = 1, l, r, k; i <= q; i++) {
    scanf("%d %d %d", &l, &r, &k);
    Q[k].push_back((query){l, r, i});
  }
  for (int i = tmp[0] = 1; i <= 200000; i++) {
    tmp[i] = qp(i, P - 2);
  }
  for (int k = 0; k <= 100000; k++)
    if (!Q[k].empty()) {
      int SZ = n / sqrt(Q[k].size()) * 2 + 1;
      for (int i = mul[0] = 1; i <= n; i++) {
        long long t = (1LL * k * m + i) % P;
        mul[i] = 1LL * mul[i - 1] * t % P;
      }
      for (int i = 1; i <= m; i++) {
        int all = cnt[i] + k;
        val[i].clear(), val[i].push_back(1);
        inv[i].clear(), inv[i].push_back(1);
        for (int j = 1; j <= cnt[i]; j++) {
          int t = all - j + 1;
          val[i].push_back(1LL * val[i].back() * t % P);
          inv[i].push_back(1LL * inv[i].back() * tmp[t] % P);
        }
      }
      sort(Q[k].begin(), Q[k].end(), [&](query a, query b) {
        int t1 = (a.l - 1) / SZ, t2 = (b.l - 1) / SZ;
        if (t1 ^ t2) return t1 < t2;
        if (t1 & 1) return a.r > b.r;
        return a.r < b.r;
      });
      memset(occ, 0, sizeof(occ));
      int l = 1, r = 0, cur = 1, cur_pw = 0;
      auto ins = [&](int x, int y) {
        cur = 1LL * cur * inv[a[x]][occ[a[x]]] % P;
        occ[a[x]] += y;
        cur = 1LL * cur * val[a[x]][occ[a[x]]] % P;
      };
      for (auto p : Q[k]) {
        while (l > p.l) ins(--l, 1);
        while (r < p.r) ins(++r, 1);
        while (l < p.l) ins(l++, -1);
        while (r > p.r) ins(r--, -1);
        ans[p.id] = 1LL * cur * mul[n - (p.r - p.l + 1)] % P;
      }
    }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}