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 M = 1100000;
const int maxn = 210000;
const long long P = 998244353;
const long long mod = 998244353;
const long long g = 3;
int N, na, nb, n, k, m, q;
long long a[maxn * 2], b[maxn * 2], d[maxn * 2], W[2][maxn * 2], rev[maxn * 2],
    f[maxn * 2], tmp1[maxn * 2], tmp2[maxn * 2];
long long inv[410000];
long long ff[410000];
long long pow_mod(long long a, int b) {
  long long c = 1;
  for (; b; b >>= 1, a = a * a % P)
    if (b & 1) c = c * a % P;
  return c;
}
void NTT(long long* a, int f, int N) {
  for (int i = 0; i < N; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1; i < N; i <<= 1)
    for (int j = 0, t = N / (i << 1); j < N; j += i << 1)
      for (int k = 0, l = 0, x, y; k < i; k++, l += t)
        x = (long long)W[f][l] * a[j + k + i] % P, y = a[j + k],
        a[j + k] = (y + x) % P, a[j + k + i] = (y - x + P) % P;
  if (f)
    for (int i = 0, x = pow_mod(N, P - 2); i < N; i++)
      a[i] = (long long)a[i] * x % P;
}
void change(int N) {
  for (int i = 0; i < N; i++) {
    int x = i, y = 0;
    for (int k = 1; k < N; x >>= 1, k <<= 1) (y <<= 1) |= x & 1;
    rev[i] = y;
  }
}
void init(int N) {
  W[0][0] = W[1][0] = 1;
  for (int i = 1, x = pow_mod(g, (P - 1) / N), y = pow_mod(x, P - 2); i < N;
       i++)
    W[0][i] = (long long)x * W[0][i - 1] % P,
    W[1][i] = (long long)y * W[1][i - 1] % P;
  change(N);
}
int len;
int main() {
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  scanf("%d", &q);
  N = 1;
  while (N < 2 * (n + 1)) N *= 2, len++;
  while (q--) {
    for (int i = 1; i <= n; i++) b[i] = a[i];
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int i, d;
      scanf("%d %d %d", &m, &i, &d);
      b[i] = d;
    } else {
      int l, r, d;
      scanf("%d %d %d %d", &m, &l, &r, &d);
      for (int i = l; i <= r; i++) b[i] = (b[i] + d) % mod;
    }
    for (int i = 1; i <= n; i++) b[i] = (m + mod - b[i]) % mod;
    for (int i = 0; i < N; i++) d[i] = (i & 1) ^ 1;
    for (int i = 0; i < n; i++) d[2 * i + 1] = b[i + 1];
    for (int i = 2; i <= len; i++) {
      int tN = 1 << i;
      init(tN);
      for (int j = 0; j < N; j += tN) {
        for (int k = j; k < j + tN / 2; k++)
          tmp1[k - j] = d[k], tmp2[k - j] = d[k + tN / 2];
        for (int k = tN / 2; k < tN; k++) tmp1[k] = tmp2[k] = 0;
        NTT(tmp1, 0, tN);
        NTT(tmp2, 0, tN);
        for (int k = 0; k < tN; k++) tmp1[k] = tmp1[k] * tmp2[k] % mod;
        NTT(tmp1, 1, tN);
        for (int k = 0; k < tN; k++) d[j + k] = tmp1[k];
      }
    }
    cout << d[k] << endl;
  }
  return 0;
}