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;
using namespace chrono;
const int infinity = (int)1e9 + 42;
const int64_t llInfinity = (int64_t)1e18 + 256;
const int module = 998244353;
const long double eps = 1e-8;
mt19937_64 randGen(system_clock().now().time_since_epoch().count());
inline void raiseError(string errorCode) {
  cerr << "Error : " << errorCode << endl;
  exit(42);
}
inline int64_t mulMod(int64_t a, int64_t b) { return (a * b) % module; }
inline int64_t powMod(int64_t a, int64_t n) {
  if (n == 1) {
    return a;
  }
  int64_t res = 1;
  while (n) {
    if (n & 1) {
      res *= a;
      res %= module;
    }
    a *= a;
    a %= module;
    n >>= 1;
  }
  return res;
}
inline int64_t rev(int64_t a) { return powMod(a, module - 2); }
struct FreqCounter {
  vector<int> freq;
  vector<int> kfr;
  inline void change(int x, int delta) {
    kfr[freq[x]]--;
    freq[x] += delta;
    kfr[freq[x]]++;
  }
  inline void add(int x) { change(x, +1); }
  inline void del(int x) { change(x, -1); }
  FreqCounter(int n, int m) : freq(m), kfr(n + 1) { kfr[0] = m; }
};
struct Query {
  int l, r, k, idx;
};
const int block = 200;
const int mfact = 300000;
int f[mfact], rf[mfact];
inline bool kCmp(const Query &a, const Query &b) {
  return make_pair(a.k, a.l - a.r) < make_pair(b.k, b.l - b.r);
}
inline bool moCmp(const Query &a, const Query &b) {
  return make_pair(a.l / block, a.r) < make_pair(b.l / block, b.r);
}
signed main() {
  ios_base::sync_with_stdio(false);
  f[0] = rf[0] = 1;
  for (int i = 1; i < mfact; i++) {
    f[i] = mulMod(f[i - 1], i);
    rf[i] = rev(f[i]);
  }
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> e(n);
  vector<int> knt(m);
  for (int i = 0; i < n; i++) {
    cin >> e[i];
    e[i]--;
    knt[e[i]]++;
  }
  const int hlim = min(n + 1, 200);
  vector<int> heavies;
  for (int i = 0; i < m; i++) {
    if (knt[i] >= hlim) {
      heavies.push_back(i);
    }
  }
  vector<Query> qry(q);
  vector<int> res(q);
  int qktr = 0;
  for (Query &q : qry) {
    cin >> q.l >> q.r >> q.k;
    q.l--;
    q.r--;
    q.idx = qktr++;
  }
  sort(qry.begin(), qry.end(), kCmp);
  for (int r = 0; r < q;) {
    int l = r;
    int qk = qry[l].k;
    while (r < q && qk == qry[r].k) {
      r++;
    }
    int64_t bans = 1;
    for (int i = 0; i < m; i++) {
      bans = mulMod(bans, f[qk + knt[i]]);
    }
    int ptr = n, fct = 1;
    for (int i = l; i < r; i++) {
      int len = qry[i].r - qry[i].l + 1;
      while (ptr > len) {
        ptr--;
        fct = mulMod(fct, ((int64_t)m * qk + n - ptr) % module);
      }
      res[qry[i].idx] = mulMod(bans, fct);
    }
  }
  sort(qry.begin(), qry.end(), moCmp);
  int l = 0, r = -1;
  FreqCounter counter(n, m);
  for (int i = 0; i < n; i++) {
    counter.add(e[i]);
  }
  for (Query q : qry) {
    while (q.l < l) {
      counter.del(e[--l]);
    }
    while (q.r > r) {
      counter.del(e[++r]);
    }
    while (q.l > l) {
      counter.add(e[l++]);
    }
    while (q.r < r) {
      counter.add(e[r--]);
    }
    int64_t ans = 1;
    for (int i = 0; i < hlim; i++) {
      if (counter.kfr[i]) {
        ans = mulMod(ans, powMod(rf[q.k + i], counter.kfr[i]));
      }
    }
    for (int i : heavies) {
      int freq = counter.freq[i];
      if (freq >= hlim) {
        ans = mulMod(ans, rf[q.k + freq]);
      }
    }
    res[q.idx] = mulMod(res[q.idx], ans);
  }
  for (int i = 0; i < q; i++) {
    cout << res[i] << "\n";
  }
  return 0;
}