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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int invnum[100005 * 2];
void initfac(int n) {
  invnum[0] = 1;
  for (int(i) = 1; (i) <= (int)(n + n); (i)++) {
    invnum[i] = fastPow(i, 998244353 - 2);
  }
}
int e[100005], SZ;
struct mb {
  int l, r, id, k;
  void rd(int i) {
    scanf("%d%d%d", &l, &r, &k);
    id = i;
  }
  bool operator<(const mb& o) const {
    if (k != o.k) return k < o.k;
    if (l / SZ == o.l / SZ) return l / SZ & 1 ? r > o.r : r < o.r;
    return l / SZ < o.l / SZ;
  }
} qs[100005];
int tt[100005], cnt[100005], tans, Ans[100005], ak[100005], clcnt;
void add(int num) {
  tans = (long long)tans * (tt[num] - cnt[num]) % 998244353;
  cnt[num]++;
}
void mov(int num) {
  cnt[num]--;
  tans = (long long)tans * invnum[tt[num] - cnt[num]] % 998244353;
}
void cal(int tl, int tr, int K) {
  memset(tt, 0, sizeof tt);
  memset(cnt, 0, sizeof cnt);
  for (int(i) = 1; (i) <= (int)(n); (i)++) tt[e[i]]++;
  for (int(i) = 1; (i) <= (int)(m); (i)++) tt[i] += K;
  long long S = (long long)K * m + n;
  ak[n] = 1;
  for (int i = n - 1; i; i--) {
    ak[i] = (long long)ak[i + 1] * ((S - i) % 998244353) % 998244353;
  }
  tans = 1;
  int L = 1, R = 0;
  for (int i = tl; i <= tr; i++) {
    while (R < qs[i].r) add(e[++R]);
    while (qs[i].r < R) mov(e[R--]);
    while (L < qs[i].l) mov(e[L++]);
    while (qs[i].l < L) add(e[--L]);
    Ans[qs[i].id] = (long long)tans * ak[qs[i].r - qs[i].l + 1] % 998244353;
  }
}
void fmain(int ID) {
  int q;
  scanf("%d%d%d", &n, &m, &q);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", e + i);
  for (int(i) = 1; (i) <= (int)(q); (i)++) {
    qs[i].rd(i);
  }
  SZ = sqrt(n);
  initfac(100005 - 1);
  sort(qs + 1, qs + 1 + q);
  for (int i = 1, j; i <= q; i = j) {
    for (j = i; j <= q && qs[i].k == qs[j].k; j++)
      ;
    cal(i, j - 1, qs[i].k);
  }
  for (int(i) = 1; (i) <= (int)(q); (i)++) printf("%d\n", Ans[i]);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}