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 std::vector;
const int N = 80005, mu = 998244353, inv3 = (mu + 1) / 3;
int k, q, Q, c[N], opt, L, R, y, x, n, m, a[N], b[N], lim, l, r[N];
void reduce(int &x) { x += x >> 31 & mu; }
void init(int n) {
  lim = 1, l = 0;
  while (lim <= n) lim <<= 1, l++;
  for (int i = 0; i < lim; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
}
int ksm(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = (long long)x * x % mu)
    if (y & 1) ans = (long long)ans * x % mu;
  return ans;
}
void fft(vector<int> &a, int flag) {
  a.resize(lim);
  for (int i = 0; i < lim; i++)
    if (i < r[i]) std::swap(a[i], a[r[i]]);
  for (int mid = 1; mid < lim; mid <<= 1) {
    int w = ksm((flag == 1 ? 3 : inv3), (mu - 1) / (mid << 1));
    for (int j = 0; j < lim; j += (mid << 1))
      for (int k = 0, ww = 1; k < mid; k++, ww = (long long)ww * w % mu) {
        int x = a[j + k], y = (long long)ww * a[j + mid + k] % mu;
        reduce(a[j + k] = x + y - mu);
        reduce(a[j + mid + k] = x - y);
      }
  }
  if (flag == -1) {
    int inv = ksm(lim, mu - 2);
    for (int i = 0; i < lim; i++) a[i] = (long long)a[i] * inv % mu;
  }
}
vector<int> work(int L, int R) {
  int mid = (L + R) >> 1;
  if (L == R) {
    return vector<int>{1, b[L]};
  }
  vector<int> pl = work(L, mid), pr = work(mid + 1, R);
  int n = pl.size(), m = pr.size();
  init(n + m);
  fft(pl, 1), fft(pr, 1);
  for (int i = 0; i < lim; i++) pl[i] = (long long)pl[i] * pr[i] % mu;
  fft(pl, -1);
  pl.resize(n + m + 1);
  return pl;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &opt, &q);
    for (int i = 1; i <= n; i++) b[i] = c[i];
    if (opt == 1) {
      scanf("%d%d", &x, &y);
      b[x] = y;
    } else {
      scanf("%d%d%d", &L, &R, &y);
      for (int i = L; i <= R; i++) reduce(b[i] += y - mu);
    }
    for (int i = 1; i <= n; i++) reduce(b[i] = q - b[i]);
    printf("%d\n", work(1, n)[k]);
  }
}