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 maxn = 100005, maxm = 100005, INF = 1000000000;
inline int read() {
  int out = 0, flag = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) {
    out = (out << 3) + (out << 1) + c - 48;
    c = getchar();
  }
  return out * flag;
}
const int G = 3, P = 998244353;
inline int qpow(int a, long long int b) {
  int re = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) re = 1ll * re * a % P;
  return re;
}
int R[maxn];
inline void NTT(int* a, int n, int f) {
  for (int i = 0; i < n; i++)
    if (i < R[i]) swap(a[i], a[R[i]]);
  for (int i = 1; i < n; i <<= 1) {
    int gn = qpow(G, (P - 1) / (i << 1));
    for (int j = 0; j < n; j += (i << 1)) {
      int g = 1, x, y;
      for (int k = 0; k < i; k++, g = 1ll * g * gn % P) {
        x = a[j + k], y = 1ll * g * a[j + k + i] % P;
        a[j + k] = (x + y) % P, a[j + k + i] = ((x - y) % P + P) % P;
      }
    }
  }
  if (f == 1) return;
  int nv = qpow(n, P - 2);
  reverse(a + 1, a + n);
  for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int A[30][maxn], deg[maxn], cnt;
int n, w[maxn];
int K, s[maxn];
void solve(int l, int r) {
  if (l == r) {
    ++cnt;
    deg[cnt] = 1;
    A[cnt][0] = 1;
    A[cnt][1] = w[l];
    return;
  }
  int mid = l + r >> 1;
  solve(l, mid);
  solve(mid + 1, r);
  int n = 1, L = 0, a = cnt - 1, b = cnt, m = deg[a] + deg[b];
  while (n <= m) n <<= 1, L++;
  for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
  for (int i = deg[a] + 1; i < n; i++) A[a][i] = 0;
  for (int i = deg[b] + 1; i < n; i++) A[b][i] = 0;
  NTT(A[a], n, 1);
  NTT(A[b], n, 1);
  for (int i = 0; i < n; i++) A[a][i] = 1ll * A[a][i] * A[b][i] % P;
  NTT(A[a], n, -1);
  cnt--;
  deg[cnt] = m;
}
int main() {
  n = read();
  K = read();
  for (int i = 1; i <= (n); i++) s[i] = read();
  int M = read(), q, L, R, d;
  while (M--) {
    int t = read();
    cnt = 0;
    if (t & 1) {
      q = read();
      L = read();
      d = read();
      if (n < K)
        puts("0");
      else {
        for (int i = 1; i <= n; i++) w[i] = (q - s[i] + P) % P;
        w[L] = (q - d + P) % P;
        solve(1, n);
        printf("%d\n", A[1][K]);
      }
    } else {
      q = read();
      L = read();
      R = read();
      d = read();
      if (n < K)
        puts("0");
      else {
        for (int i = 1; i <= n; i++) w[i] = (q - s[i] + P) % P;
        for (int i = L; i <= R; i++) w[i] = (q - s[i] - d + P) % P;
        solve(1, n);
        printf("%d\n", A[1][K]);
      }
    }
  }
  return 0;
}
