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 MOD = 998244353;
int N, K;
int A[20010], B[20010];
int f[20010];
int solve() {
  for (int i = 1; i <= N; i++) {
    B[i] = (B[i] % MOD + MOD) % MOD;
  }
  for (int i = 0; i <= N; i++) {
    f[i] = 0;
  }
  f[0] = 1;
  for (int i = 1; i <= N; i++) {
    int down = max(K - (N - i), 1);
    for (int k = min(K, i); k >= down; k--) {
      f[k] = (f[k] + 1LL * f[k - 1] * B[i]) % MOD;
    }
  }
  return f[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 tp, q, L, R, i, d;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d%d%d", &q, &i, &d);
      for (int j = 1; j <= N; j++) {
        B[j] = q - A[j];
      }
      B[i] = q - d;
    } else {
      scanf("%d%d%d%d", &q, &L, &R, &d);
      for (int j = 1; j <= N; j++) {
        if (j >= L && j <= R) {
          B[j] = q - (A[j] + d);
        } else {
          B[j] = q - A[j];
        }
      }
    }
    printf("%d\n", solve());
  }
}
