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;
template <typename U, typename V>
void Min(U &a, const V &b) {
  if (a > b) a = b;
}
template <typename U, typename V>
void Max(U &a, const V &b) {
  if (a < b) a = b;
}
template <typename U, typename V>
void add(U &a, const V &b) {
  a = (a + b) % 998244353;
}
template <typename U>
U gcd(U a, U b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a >= b)
    return gcd(a - b, b);
  else
    return gcd(a, b - a);
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 998244353;
    a = 1LL * a * a % 998244353;
    b >>= 1;
  }
  return ans;
}
int a[20005], b[20005], dp[20005], f[20005];
int main() {
  int T, m, n, i, j, k, K;
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  long long q;
  while (m--) {
    scanf("%d%lld", &k, &q);
    for (int i = 0; i < n; i++) b[i] = a[i];
    if (k == 1) {
      scanf("%d%d", &i, &j);
      b[i - 1] = j;
    } else {
      scanf("%d%d%d", &i, &j, &k);
      i--;
      for (int l = i; l < j; l++) {
        b[l] = (b[l] + k) % 998244353;
      }
    }
    for (int i = 0; i < n; i++) {
      b[i] = (q - b[i] + 998244353) % 998244353;
    }
    for (int j = 0; j < n + 1; j++) dp[j] = 0;
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
      int l = n - i < K ? i + K - n : 0;
      int r = i + 1 >= K ? K - 1 : i;
      for (int j = r + 1 - 1; j >= l; j--) {
        dp[j + 1] = (dp[j + 1] + 1LL * dp[j] * b[i]) % 998244353;
      }
    }
    printf("%d\n", dp[K]);
  }
}
