Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read(register long long a = 0, register char ch = getchar()) {
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    a = (a << 3) + (a << 1) + (ch ^ 48), ch = getchar();
  return a;
}
const int mod = 998244353;
inline int mo(const int x) { return x >= mod ? x - mod : x; }
inline int qpow(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
int v[33000];
inline void ntt(int *A, const int bin, const int opt) {
  for (int i = 0; i < bin; i++) v[i] = v[i >> 1] >> 1 | (i & 1 ? bin >> 1 : 0);
  for (int i = 0; i < bin; i++)
    if (i < v[i]) swap(A[i], A[v[i]]);
  for (int i = 1; i < bin; i <<= 1) {
    const int wn = qpow(3, opt * (mod - 1) / (i << 1) + mod - 1);
    for (int j = 0, I = i << 1; j < bin; j += I) {
      long long w = 1;
      for (int k = 0; k < i; k++, w = w * wn % mod) {
        const int x = A[j + k], y = A[j + k + i] * w % mod;
        A[j + k] = mo(x + y);
        A[j + k + i] = mo(x - y + mod);
      }
    }
  }
  if (opt == -1)
    for (int i = 0, iv = qpow(bin, mod - 2); i < bin; i++)
      A[i] = 1ll * A[i] * iv % mod;
}
int a[20010], A[33000], B[33000], tmp[20010];
vector<int> f[80010];
inline void rs(vector<int> &T, const int bin) {
  T.resize(bin);
  for (int i = 0; i < bin; i++) T[i] = 0;
}
void prod(const int k, const int l, const int r, int bin = 1) {
  if (l == r) {
    rs(f[k], 2);
    f[k][0] = 1;
    f[k][1] = tmp[l];
    return;
  }
  const int mid = l + r >> 1;
  prod(k << 1, l, mid);
  prod(k << 1 | 1, mid + 1, r);
  const int len = f[k << 1].size() + f[k << 1 | 1].size() - 1;
  rs(f[k], len);
  while (bin < len) bin <<= 1;
  for (int i = 0; i < bin; i++) A[i] = B[i] = 0;
  for (int i = 0, I = f[k << 1].size(); i < I; i++) A[i] = f[k << 1][i];
  for (int i = 0, I = f[k << 1 | 1].size(); i < I; i++) B[i] = f[k << 1 | 1][i];
  ntt(A, bin, 1);
  ntt(B, bin, 1);
  for (int i = 0; i < bin; i++) A[i] = 1ll * A[i] * B[i] % mod;
  ntt(A, bin, -1);
  for (int i = 0; i < len; i++) f[k][i] = A[i];
}
int main() {
  const int n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int Q = read();
  while (Q--) {
    const int opt = read();
    if (opt == 1) {
      const int q = read(), p = read(), d = read();
      for (int i = 1; i <= n; i++)
        if (i == p)
          tmp[i] = (q - d + mod) % mod;
        else
          tmp[i] = (q - a[i] + mod) % mod;
    } else {
      const int q = read(), L = read(), R = read(), d = read();
      for (int i = 1; i <= n; i++)
        if (i >= L && i <= R)
          tmp[i] = ((q - a[i] - d) % mod + mod) % mod;
        else
          tmp[i] = (q - a[i] + mod) % mod;
    }
    prod(1, 1, n);
    printf("%d\n", f[1][k]);
  }
}