While roaming the mystic areas of Stonefalls, in order to drop legendary loot, an adventurer was given a quest as follows. He was given an array A = {a_1,a_2,...,a_N } of length N, and a number K.

Define array B as B(q, A) =  { q-a_1, q-a_2, ..., q-a_N }. Define function F as F(B,K) being sum of products of all K-tuples of elements in array B. For example, if the array B is [2,3,4,5], and with K=3, sum of products of all 3-tuples is $$$F(B, 3) = 2*3*4+2*3*5+3*4*5+2*4*5$$$

He was then given a number Q, number of queries of two types: 

  * Type 1: Given q, i, and d calculate F(B(q, A), K) where we make change to initial array as A[i] = d. 
  * Type 2: Given q, L, R, and d calculate F(B(q, A), K) where we make change to initial array as A[i] = A[i] + d for all i in range [L, R] inclusive. 



All changes are temporarily made to initial array, and don't propagate to following queries. Help the adventurer calculate the answer to a quest, and finally get that loot!

Input

In the first two lines, numbers N (1 ≤ N ≤ 2*10^4) and K (1 ≤ K ≤ N), the length of initial array A, and tuple size, followed by a_1,a_2,a_3,…,a_N (0 ≤ a_i ≤ 10^9) , elements of array A, in the next line. Then follows number Q (Q ≤ 10), number of queries. In the next Q lines come queries of the form: 

  * 1 q i d, for type 1, 
  * 2 q L R d, for type 2, 



as explained above (0 ≤ q, d ≤ 10^9, 1 ≤ i,L,R ≤ N)

Output

Print Q lines, the answers to queries, modulo 998244353.

Example

Input


5
2
1 2 3 4 5
3
1 6 1 1
1 6 5 2
2 6 2 3 1


Output


85
127
63

Note

In the first query array A = [1, 2, 3, 4, 5], B = [5, 4, 3, 2, 1], sum of products of 2-tuples = 85.

In second query array A = [1, 2, 3, 4, 2], B = [5, 4, 3, 2, 4], sum of products of 2-tuples = 127

In third query array A = [1, 3, 4, 4, 5], B = [5, 3, 2, 2, 1], sum of products of 2-tuples = 63
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void cetak(T t) {
  cout << t << ')' << endl;
}
template <typename T, typename... V>
void cetak(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cerr << ", ";
  cetak(v...);
}
const int MOD = 1e9 + 7;
const int mx = 1e5 + 10;
struct NTT {
 public:
  NTT(const int _mod) : mod(_mod) {
    proot = find_primitive_root();
    int totient = mod - 1;
    max_len = 1;
    while ((totient & max_len) == 0) max_len <<= 1;
  }
  void fft(vector<int> &a, bool invert) {
    int n = (int)a.size();
    if (n & (n - 1)) {
      int pw = 1;
      while (pw < n) pw <<= 1;
      n = pw;
      a.resize(n, 0);
    }
    assert(n <= max_len);
    for (int i = 1, j = 0; i < n; ++i) {
      int bit = n >> 1;
      for (; j >= bit; bit >>= 1) j -= bit;
      j += bit;
      if (i < j) swap(a[i], a[j]);
    }
    int root = powmod(proot, (mod - 1) / n);
    if (invert) root = inverse(root);
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = root;
      for (int i = len; i < n; i <<= 1) wlen = int(wlen * 1ll * wlen % mod);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; ++j) {
          int u = a[i + j], v = int(a[i + j + len / 2] * 1ll * w % mod);
          a[i + j] = u + v < mod ? u + v : u + v - mod;
          a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
          w = int(w * 1ll * wlen % mod);
        }
      }
    }
    if (invert) {
      int nrev = inverse(n);
      for (int i = 0; i < n; ++i) a[i] = int(a[i] * 1ll * nrev % mod);
    }
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int len = a.size() + b.size() - 1;
    int pw = 1;
    while (pw < len) pw <<= 1;
    len = pw;
    a.resize(len, 0);
    b.resize(len, 0);
    fft(a, false);
    fft(b, false);
    for (int i = 0; i < len; ++i) a[i] = 1LL * a[i] * b[i] % mod;
    fft(a, true);
    while (len > 0 && a[len - 1] == 0) --len;
    a.resize(len);
    return a;
  }
  long long powmod(long long b, long long p) {
    long long r = 1;
    for (; p; p >>= 1, b = b * b % mod)
      if (p & 1) r = r * b % mod;
    return r;
  }
  long long inverse(int x) { return powmod(x, mod - 2); }

 private:
  const int mod;
  int proot;
  int max_len = 1;
  int find_primitive_root() {
    int totient = mod - 1;
    vector<int> checks;
    int tmp = totient;
    for (int p = 2; p * p <= tmp; ++p) {
      if (tmp % p) continue;
      checks.push_back(totient / p);
      while ((tmp % p) == 0) tmp /= p;
    }
    if (tmp > 1) checks.push_back(totient / tmp);
    for (int r = 2;; ++r) {
      bool ok = 1;
      for (int first : checks) {
        if (powmod(r, first) == 1) {
          ok = 0;
          break;
        }
      }
      if (ok) return r;
    }
    assert(false);
  }
};
const int mod = 998244353;
NTT ntt(mod);
vector<int> kali(vector<vector<int>> &ve, int l, int r) {
  if (l == r) return ve[l];
  int mid = (l + r) >> 1;
  auto kiri = kali(ve, l, mid);
  auto kanan = kali(ve, mid + 1, r);
  return ntt.multiply(kiri, kanan);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int q, x, d;
      cin >> q >> x >> d;
      x -= 1;
      auto b = a;
      b[x] = d;
      for (int i = 0; i < n; i++) b[i] = q - b[i];
      for (int &i : b) {
        i %= mod;
        if (i < 0) i += mod;
      }
      vector<vector<int>> ve(n, vector<int>(2));
      for (int i = 0; i < n; i++) {
        ve[i][0] = 1;
        ve[i][1] = b[i];
      }
      auto res = kali(ve, 0, n - 1);
      if (k >= (int)(res.size()))
        cout << 0 << '\n';
      else
        cout << res[k] << '\n';
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      l -= 1, r -= 1;
      auto b = a;
      for (int i = l; i <= r; i++) b[i] = a[i] + d;
      for (int &i : b) i = q - i;
      for (int &i : b) {
        i %= mod;
        if (i < 0) i += mod;
      }
      vector<vector<int>> ve(n, vector<int>(2));
      for (int i = 0; i < n; i++) {
        ve[i][0] = 1;
        ve[i][1] = b[i];
      }
      auto res = kali(ve, 0, n - 1);
      if (k >= (int)(res.size()))
        cout << 0 << '\n';
      else
        cout << res[k] << '\n';
    }
  }
}
