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")
#pragma GCC optimize("inline")
using namespace std;
const int N = 200005, MOD = 998244353, BLO = 987;
int n, m, Q, k, res;
int fac[N], ifac[N], inv[N], prd[N];
int a[N], t[N], c[N], ans[N];
struct Que {
  int l, r, k, id;
  inline friend bool operator<(Que a, Que b) {
    if (a.k != b.k) return a.k < b.k;
    return (a.l / BLO != b.l / BLO) ? (a.l < b.l) : (a.r < b.r);
  }
} q[N];
int Down(int x, int y) {
  if (x < y) throw;
  return (long long)fac[x] * ifac[x - y] % MOD;
}
void Inc(int x) {
  if (!x) return;
  ++c[a[x]];
  res = (long long)res * (t[a[x]] + k - c[a[x]] + 1) % MOD;
}
void Dec(int x) {
  if (!x) return;
  res = (long long)res * inv[t[a[x]] + k - c[a[x]] + 1] % MOD;
  --c[a[x]];
}
void Solve(int ql, int qr) {
  int mk = (long long)k * m % MOD;
  prd[n] = 1;
  for (int i = n - 1; ~i; --i) {
    prd[i] = (long long)prd[i + 1] * (mk + n - i) % MOD;
  }
  res = 1;
  int l = 1, r = 0, L, R;
  for (int i = ql; i <= qr; ++i) {
    L = q[i].l;
    R = q[i].r;
    while (r < R) Inc(++r);
    while (l > L) Inc(--l);
    while (l < L) Dec(l++);
    while (r > R) Dec(r--);
    ans[q[i].id] = (long long)res * prd[r - l + 1] % MOD;
  }
  for (int i = l; i <= r; ++i) --c[a[i]];
}
int main() {
  fac[0] = fac[1] = ifac[0] = ifac[1] = inv[1] = 1;
  for (int i = 2; i < N; ++i) {
    inv[i] = MOD - (long long)(MOD / i) * inv[MOD % i] % MOD;
    fac[i] = (long long)fac[i - 1] * i % MOD;
    ifac[i] = (long long)ifac[i - 1] * inv[i] % MOD;
  }
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ++t[a[i]];
  }
  for (int i = 1; i <= Q; ++i) {
    scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].k);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + Q);
  for (int i = 1, j; i <= Q; i = j + 1) {
    for (j = i; j < Q && q[j + 1].k == q[j].k; ++j)
      ;
    k = q[i].k;
    Solve(i, j);
  }
  for (int i = 1; i <= Q; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}