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 max_n = 20222, mod = 998244353;
int n, k, q, A[max_n], a[max_n];
int dp[max_n];
int solve() {
  int q1 = 0, q2 = 1;
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = min(i + 1, k); j >= max(1, k - (n - i - 1)); --j) {
      dp[j] = (dp[j] + 1LL * dp[j - 1] * a[i]) % mod;
    }
  }
  return dp[k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
  }
  scanf("%d", &q);
  while (q--) {
    copy(A, A + n, a);
    int tp, x;
    scanf("%d%d", &tp, &x);
    if (tp == 1) {
      int pos, d;
      scanf("%d%d", &pos, &d);
      --pos;
      a[pos] = d;
    } else {
      int l, r, d;
      scanf("%d%d%d", &l, &r, &d);
      --l;
      for (int i = l; i < r; ++i) {
        a[i] += d;
      }
    }
    for (int i = 0; i < n; ++i) {
      a[i] = ((x - a[i]) % mod + mod) % mod;
    }
    printf("%d\n", solve());
  }
  return 0;
}