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;
const int maxn = 20010, mod = 998244353, g = 3;
inline int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) res = 1ll * res * a % mod;
  return res;
}
int rev[maxn << 2], ntt_a[maxn << 2], ntt_b[maxn << 2];
inline void ntt(int *f, int n, int opt) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[rev[i]]);
  for (int p = 2; p <= n; p <<= 1) {
    int len = p >> 1, tmp = ksm(g, (mod - 1) / p);
    if (opt == -1) tmp = ksm(tmp, mod - 2);
    for (int k = 0; k < n; k += p) {
      int buf = 1;
      for (int l = k; l < k + len; l++) {
        int tt = 1ll * f[l + len] * buf % mod;
        f[l + len] = f[l] - tt + mod;
        if (f[l + len] >= mod) f[l + len] -= mod;
        f[l] = f[l] + tt;
        if (f[l] >= mod) f[l] -= mod;
        buf = 1ll * buf * tmp % mod;
      }
    }
  }
  if (opt == -1)
    for (int i = 0, t = ksm(n, mod - 2); i < n; i++)
      f[i] = 1ll * f[i] * t % mod;
}
inline vector<int> Mul(const vector<int> &a, const vector<int> &b) {
  int L = 1;
  for (; L <= (int)a.size() + (int)b.size(); L <<= 1)
    ;
  for (int i = 0; i < L; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (L >> 1) : 0);
  for (int i = 0; i < (int)a.size(); i++) ntt_a[i] = a[i];
  for (int i = 0; i < (int)b.size(); i++) ntt_b[i] = b[i];
  for (int i = a.size(); i < L; i++) ntt_a[i] = 0;
  for (int i = b.size(); i < L; i++) ntt_b[i] = 0;
  ntt(ntt_a, L, 1);
  ntt(ntt_b, L, 1);
  for (int i = 0; i < L; i++) ntt_a[i] = 1ll * ntt_a[i] * ntt_b[i] % mod;
  ntt(ntt_a, L, -1);
  vector<int> ans;
  while (L && ntt_a[L - 1] == 0) L--;
  for (int i = 0; i < L; i++) ans.push_back(ntt_a[i]);
  return ans;
}
int n, k, a[maxn], b[maxn];
vector<int> poly[maxn];
inline vector<int> solve(int l, int r) {
  if (l == r) return poly[l];
  int mid = (l + r) >> 1;
  return Mul(solve(l, mid), solve(mid + 1, r));
}
inline int getans() {
  for (int i = 1; i <= n; i++)
    poly[i].clear(), poly[i].push_back(1), poly[i].push_back(b[i]);
  vector<int> t = solve(1, n);
  return (k >= (int)t.size()) ? 0 : t[k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int opt, q, x, y, d;
    scanf("%d", &opt);
    for (int i = 1; i <= n; i++) b[i] = a[i];
    if (opt == 1) {
      scanf("%d%d%d", &q, &x, &d);
      b[x] = d;
      for (int i = 1; i <= n; i++) b[i] = (q - b[i] + mod) % mod;
      printf("%d\n", getans());
    }
    if (opt == 2) {
      scanf("%d%d%d%d", &q, &x, &y, &d);
      for (int i = x; i <= y; i++) b[i] = (b[i] + d) % mod;
      for (int i = 1; i <= n; i++) b[i] = (q - b[i] + mod) % mod;
      printf("%d\n", getans());
    }
  }
  return 0;
}