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 = 1e5;
const int mod = 998244353, modg = 3, modi = 332748118;
long long a[maxn], b[maxn];
long long pow_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
class nubmer {
 public:
  int rev[maxn], len, pw;
  void init(int n) {
    len = 1;
    pw = 0;
    while (len <= n) len <<= 1, pw++;
    for (int i = 0; i <= len - 1; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (pw - 1));
  }
  void transform(long long* a, int flag) {
    for (int i = 0; i <= len - 1; i++)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
      long long wn = pow_mod(flag == 1 ? modg : modi, (mod - 1) / (mid << 1));
      for (int r = mid << 1, j = 0; j < len; j += r) {
        long long t = 1;
        for (int k = 0; k < mid; k++, t = t * wn % mod) {
          long long x = a[j + k], y = t * a[mid + j + k];
          a[j + k] = (x + y) % mod, a[j + k + mid] = (x - y + mod) % mod;
        }
      }
    }
    if (flag == -1) {
      long long inv = pow_mod(len, mod - 2);
      for (int i = 0; i <= len; i++) a[i] = (mod + a[i] * inv % mod) % mod;
    }
  }
} ntt;
int n, k;
long long s1[maxn], s2[maxn];
vector<long long> solve(int l, int r) {
  if (l == r) {
    return vector<long long>{1, s2[l]};
  }
  int mid = (l + r) >> 1;
  vector<long long> x = solve(l, mid), y = solve(mid + 1, r);
  ntt.init(x.size() + y.size());
  for (int i = 0; i <= ntt.len + 5; i++) a[i] = 0, b[i] = 0;
  for (int i = 0; i < x.size(); i++) a[i] = x[i];
  for (int i = 0; i < y.size(); i++) b[i] = y[i];
  ntt.transform(a, 1);
  ntt.transform(b, 1);
  for (int i = 0; i < ntt.len; i++) a[i] = a[i] * b[i] % mod;
  ntt.transform(a, -1);
  int len = x.size() + y.size();
  int mi = min(ntt.len, min(k + 1, len));
  vector<long long> ans(mi);
  for (int i = 0; i < mi; i++) {
    ans[i] = a[i];
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> s1[i];
  }
  int T;
  cin >> T;
  vector<long long> tmp;
  while (T--) {
    int flag;
    cin >> flag;
    if (flag == 1) {
      int q, i, d;
      cin >> q >> i >> d;
      for (int j = 1; j <= n; j++) s2[j] = (q - s1[j] + mod) % mod;
      s2[i] = (q - d + mod) % mod;
      tmp = solve(1, n);
      cout << tmp[k] << "\n";
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      for (int j = 1; j <= n; j++) {
        if (j >= l && j <= r)
          s2[j] = ((q - s1[j] - d) % mod + mod) % mod;
        else
          s2[j] = (q - s1[j] + mod) % mod;
      }
      tmp = solve(1, n);
      cout << tmp[k] << "\n";
    }
  }
  return 0;
}