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 M = 998244353, G = 3;
int n, m, i, r[1000005], type, q, L, R, d, Q, x, k;
long long a[100005], b[100005];
long long qpow(long long a, long long b) {
  long long s = a, ans = 1;
  while (b) {
    if (b & 1) ans = ans * s % M;
    s = s * s % M;
    b >>= 1;
  }
  return ans;
}
void NTT(long long *f, int limit, int op) {
  for (register int i = 0; i < limit; ++i)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) ? limit >> 1 : 0);
  for (register int i = 0; i < limit; ++i)
    if (i < r[i]) swap(f[i], f[r[i]]);
  for (register int i = 2; i <= limit; i <<= 1) {
    long long wn = qpow(3, (M - 1) / i * op + (M - 1) * (op == -1));
    for (register int j = 0; j < limit; j += i) {
      long long s = 1;
      for (register int k = j; k < j + (i >> 1); ++k) {
        long long a = f[k], b = f[k + (i >> 1)] * s;
        f[k] = (a + b) % M;
        f[k + (i >> 1)] = (a - b) % M;
        s = s * wn % M;
      }
    }
  }
  if (op == -1) {
    long long s = qpow(limit, M - 2);
    for (register int i = 0; i < limit; ++i) f[i] = f[i] * s % M;
  }
}
long long ans[200005];
void dfs(int l, int r) {
  if (l == r) {
    ans[0] = 1;
    ans[1] = b[l];
    return;
  }
  int mid = l + r >> 1;
  dfs(l, mid);
  int j;
  int limit = 1;
  while (limit <= r - l + 1) limit <<= 1;
  long long a[limit << 1], b[limit << 1];
  memset(a, 0, sizeof(a));
  memset(b, 0, sizeof(b));
  for (j = 0; j <= mid - l + 1; j++) a[j] = ans[j];
  dfs(mid + 1, r);
  for (j = 0; j <= r - mid; j++) b[j] = ans[j];
  NTT(a, limit, 1);
  NTT(b, limit, 1);
  for (j = 0; j < limit; j++) a[j] = a[j] * b[j] % M;
  NTT(a, limit, -1);
  for (j = 0; j < limit; j++) ans[j] = a[j];
}
int main() {
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d %d %d %d", &q, &x, &L, &R);
    for (i = 1; i <= n; i++) b[i] = x - a[i];
    if (q == 1)
      b[L] = x - R;
    else {
      scanf("%d", &d);
      for (i = L; i <= R; i++) b[i] = x - a[i] - d;
    }
    dfs(1, n);
    cout << (ans[k] % M + M) % M << endl;
  }
}
