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;
int const N = 100000, md = 998244353, SQ = 320;
int n, m, q, x[N + 1], an[N], edfr[N + 1], w, tmp[N + 1], fr[SQ], edfr2[N + 1];
pair<int, bool> fc[N + 1];
inline void ml(int &x, int y) { x = (long long)x * y % md; }
int pw(int x, int p) {
  int an = 1;
  while (p) {
    if (p & 1) an = (long long)an * x % md;
    x = (long long)x * x % md;
    p >>= 1;
  }
  return an;
}
int inv(int x) { return pw(x, md - 2); }
struct Q {
  int l, r, m, id;
  void sc(int i) {
    scanf("%d%d%d", &l, &r, &m);
    id = i;
  }
  bool operator<(Q const &o) const {
    int a = l / w, b = o.l / w;
    if (a == b) return a & 1 ? r < o.r : o.r < r;
    return a < b;
  }
} qu[N];
vector<Q> mqu[N + 1];
void fl(long long x) {
  fc[0] = make_pair(1, false);
  for (int i = 1; i < (int)(n + 1); ++i) {
    if (x % md == 0)
      fc[i] = make_pair(1, true);
    else
      fc[i] = make_pair(x % md * fc[i - 1].first % md, fc[i - 1].second);
    ++x;
  }
}
int getFc(int l, int r) {
  if (fc[r].second && !fc[l - 1].second) return 0;
  return (long long)fc[r].first * inv(fc[l - 1].first) % md;
}
void upfr(int i, int v) {
  if (i == 0) return;
  if (v == -1) --fr[edfr[i] + 1 - edfr2[i]];
  edfr2[i] += v;
  if (v == 1) ++fr[edfr[i] + 1 - edfr2[i]];
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  w = 200;
  for (int i = 1; i < (int)(n + 1); ++i) scanf("%d", x + i);
  for (int i = 1; i < (int)(n + 1); ++i) ++edfr[x[i]];
  for (int i = 0; i < (int)(q); ++i) qu[i].sc(i);
  for (int i = 0; i < (int)(q); ++i) mqu[qu[i].m].push_back(qu[i]);
  for (int i = 0; i < (int)(q); ++i) an[i] = 1;
  for (int i = 0; i < (int)(N + 1); ++i)
    if (!mqu[i].empty()) {
      fl((long long)m * i + 1);
      for (int j = 0; j < (int)(mqu[i].size()); ++j)
        ml(an[mqu[i][j].id], getFc(1, n - (mqu[i][j].r - mqu[i][j].l + 1)));
    }
  vector<vector<int> > H;
  for (int k = 1; k < (int)(m + 1); ++k)
    if (edfr[k] >= w) {
      H.push_back(vector<int>(q));
      tmp[0] = 0;
      for (int i = 1; i < (int)(n + 1); ++i) tmp[i] = tmp[i - 1] + (x[i] == k);
      for (int i = 0; i < (int)(q); ++i)
        H.back()[i] = tmp[qu[i].r] - tmp[qu[i].l - 1];
    }
  for (int i = 0; i < (int)(N + 1); ++i)
    if (!mqu[i].empty()) {
      fl(i + 1);
      int z = 0;
      for (int k = 1; k < (int)(m + 1); ++k)
        if (edfr[k] >= w) {
          for (int j = 0; j < (int)(mqu[i].size()); ++j)
            ml(an[mqu[i][j].id],
               getFc(edfr[k] - H[z][mqu[i][j].id] + 1, edfr[k]));
          ++z;
        }
    }
  sort(qu, qu + q);
  for (int i = 1; i < (int)(n + 1); ++i)
    if (edfr[x[i]] >= w) x[i] = 0;
  int l = 1, r = 0;
  for (int i = 0; i < (int)(q); ++i) {
    while (r < qu[i].r) upfr(x[++r], 1);
    while (l > qu[i].l) upfr(x[--l], 1);
    while (r > qu[i].r) upfr(x[r--], -1);
    while (l < qu[i].l) upfr(x[l++], -1);
    for (int j = 1; j < (int)(w); ++j) ml(an[qu[i].id], pw(qu[i].m + j, fr[j]));
  }
  for (int i = 0; i < (int)(q); ++i) printf("%d\n", an[i]);
}