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;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
const int mod = 998244353;
const int g = 3;
const int inv_g = 332748118;
const int maxN = 100010;
int Power(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
int inv(int x) { return Power(x, mod - 2); }
int rev[maxN];
void init(int L) {
  for (int i = 0; i < L; i++) rev[i] = rev[i / 2] / 2 + (i & 1 ? L / 2 : 0);
}
void NTT(vector<int> &A, int L, int op) {
  if (rev[1] != L / 2) init(L);
  if ((int)A.size() < L) A.resize(L);
  for (int i = 0; i < L; i++)
    if (i < rev[i]) swap(A[i], A[rev[i]]);
  for (int i = 2; i <= L; i *= 2) {
    int w_ = Power(op ? inv_g : g, (mod - 1) / i);
    for (int j = 0; j < L; j += i) {
      int w = 1;
      for (int k = 0; k < i / 2; k++) {
        int x = A[j + k], y = 1ll * w * A[j + i / 2 + k] % mod;
        A[j + k] = (x + y) % mod;
        A[j + i / 2 + k] = (x - y + mod) % mod;
        w = 1ll * w * w_ % mod;
      }
    }
  }
  if (op) {
    int tmp = inv(L);
    for (int i = 0; i < L; i++) A[i] = 1ll * A[i] * tmp % mod;
  }
}
vector<int> operator*(vector<int> A, vector<int> B) {
  int L = 1;
  while (L < (int)A.size() + (int)B.size() - 1) L <<= 1;
  NTT(A, L, 0);
  NTT(B, L, 0);
  for (int i = 0; i < L; i++) A[i] = 1ll * A[i] * B[i] % mod;
  NTT(A, L, 1);
  return A;
}
int n, K;
int A[maxN];
int Q;
int B[maxN];
vector<int> solve(int L, int R) {
  if (L == R) return {1, B[L]};
  int Mid = (L + R) / 2;
  return solve(L, Mid) * solve(Mid + 1, R);
}
vector<int> Ans;
int main() {
  n = read();
  K = read();
  for (int i = 1; i <= n; i++) A[i] = read();
  Q = read();
  while (Q--) {
    int op = read();
    if (op == 1) {
      int q = read(), I = read(), d = read();
      for (int i = 1; i <= n; i++) B[i] = (q - (i == I ? d : A[i]) + mod) % mod;
    }
    if (op == 2) {
      int q = read(), L = read(), R = read(), d = read();
      for (int i = 1; i <= n; i++) B[i] = (q - A[i] + mod) % mod;
      for (int i = L; i <= R; i++) B[i] = (B[i] - d + mod) % mod;
    }
    Ans = solve(1, n);
    printf("%d\n", Ans[K]);
  }
  return 0;
}
