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 mod = 998244353;
int w[35005], r[35005], wlen, lg[35005];
int Pow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) b & 1 && (s = 1ll * s * a % mod);
  return s;
}
void init(int n) {
  w[0] = wlen = 1;
  while (wlen <= n) wlen <<= 1;
  w[1] = Pow(3, (mod - 1) / wlen);
  for (int i = 2; i <= wlen; i++)
    w[i] = 1ll * w[i - 1] * w[1] % mod, lg[i] = lg[i >> 1] + 1;
}
int upd(int x) { return x >= mod ? x - mod : x; }
void NTT(int *a, int len, int flg) {
  for (int i = 0; i < len; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  for (int i = 2, l = 1, v; i <= len; l = i, i <<= 1)
    for (int j = 0, t = wlen / i; j < len; j += i)
      for (int k = j, o = 0; k < j + l; k++, o += t)
        a[k + l] =
            upd(a[k] - (v = 1ll * w[flg ^ 1 ? wlen - o : o] * a[k + l] % mod) +
                mod),
              a[k] = upd(a[k] + v);
  if (flg ^ 1)
    for (int i = 0, Inv = mod - (mod - 1) / len; i < len; i++)
      a[i] = 1ll * a[i] * Inv % mod;
}
int n, K, a[35005], b[35005], Q;
vector<int> Mul(const vector<int> &A, const vector<int> &B) {
  static int a[35005], b[35005];
  int n = A.size(), m = B.size(), len = 1 << (lg[n + m - 2] + 1);
  for (int i = 0; i < len; i++)
    a[i] = i < n ? A[i] : 0, b[i] = i < m ? B[i] : 0,
    r[i] = r[i >> 1] >> 1 | (i & 1 ? len >> 1 : 0);
  NTT(a, len, 1), NTT(b, len, 1);
  for (int i = 0; i < len; i++) a[i] = 1ll * a[i] * b[i] % mod;
  NTT(a, len, -1);
  return vector<int>(a, a + n + m - 1);
}
vector<int> solve(int l, int r) {
  if (l == r) return {1, b[l]};
  int mid = (l + r) >> 1;
  return Mul(solve(l, mid), solve(mid + 1, r));
}
int main() {
  scanf("%d%d", &n, &K), init(n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &Q);
  for (int op, q, l, r, d, x; Q--;) {
    memcpy(b, a, (n + 1) << 2);
    scanf("%d%d", &op, &q);
    if (op == 1)
      scanf("%d%d", &x, &d), b[x] = d;
    else {
      scanf("%d%d%d", &l, &r, &d);
      for (int i = l; i <= r; i++) b[i] = upd(b[i] + d);
    }
    for (int i = 1; i <= n; i++) b[i] = upd(q - b[i] + mod);
    printf("%d\n", solve(1, n)[K]);
  }
}