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;
const int INF = int(1e9 + 7);
const int mod = 998244353;
const int root = 15311432;
const int root_1 = 469870224;
const int root_pw = 1 << 23;
int modPow(int b, int e) {
  if (e == 0) return 1;
  int res = modPow(b, e / 2);
  if (e % 2 == 1)
    return ((1LL * b * res % mod) * res % mod) % mod;
  else
    return (1LL * res * res % mod) % mod;
}
void fft(vector<int>& a, bool invert) {
  int n = a.size();
  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]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = (int)(1LL * wlen * 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)(1LL * a[i + j + len / 2] * 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)(1LL * w * wlen % mod);
      }
    }
  }
  if (invert) {
    int n_1 = modPow(n, mod - 2);
    for (int& x : a) x = (int)(1LL * x * n_1 % mod);
  }
}
vector<int> fftMul(vector<int>& a, vector<int>& b) {
  vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (int i = 0; i < n; i++) fa[i] = (int)(1LL * fa[i] * fb[i] % mod);
  fft(fa, true);
  return fa;
}
const int tam = 40010;
int A[tam], B[tam];
vector<int> DC(int b, int e) {
  if (b == e) return {1, B[b]};
  int mid = (b + e) / 2;
  vector<int> L, R;
  L = DC(b, mid);
  R = DC(mid + 1, e);
  L = fftMul(L, R);
  return L;
}
void genB(int q, int L, int R, int d, int n) {
  for (int i = 0; i < n; i++) {
    B[i] = q - (L <= i && i <= R ? (A[i] + d) : A[i]);
    B[i] = (B[i] % mod + mod) % mod;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int q, i, d;
      cin >> q >> i >> d;
      i--;
      genB(q, i, i, (d - A[i]), n);
    } else {
      int q, L, R, d;
      cin >> q >> L >> R >> d;
      L--;
      R--;
      genB(q, L, R, d, n);
    }
    vector<int> res = DC(0, n - 1);
    cout << res[k] << '\n';
  }
  return 0;
}
