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 N = 1e5 + 500, mod = 998244353;
int K, tmp[N], tmpa[N], tmpb[N], rev[N], b[N], tw[N], w[N], limn, n, qn;
vector<int> a[N];
int qpower(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
void prework(int n) {
  limn = 1;
  while (limn <= n) limn <<= 1;
  for (int i = 1; i < limn; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (limn >> 1) : 0);
  for (int i = 1; i < limn; i <<= 1) {
    int omg = qpower(3, (mod - 1) / (i << 1));
    w[i] = 1;
    for (int j = 1; j < i; j++) w[i + j] = 1ll * w[i + j - 1] * omg % mod;
  }
}
void NTT(int a[], int l) {
  for (int i = 0; i < l; i++) tmp[rev[i] / (limn / l)] = a[i];
  for (int i = 1; i < l; i <<= 1)
    for (int j = 0; j < l; j += i << 1)
      for (int k = 0; k < i; k++) {
        int x = tmp[j + k], y = 1ll * w[i + k] * tmp[i + j + k] % mod;
        tmp[j + k] = (x + y) % mod, tmp[i + j + k] = (x - y) % mod;
      }
  for (int i = 0; i < l; i++) a[i] = tmp[i];
}
void mul(const vector<int> A, const vector<int> B, vector<int>& c) {
  int la = A.size(), lb = B.size(), n = la + lb - 1;
  int l = 1;
  while (l < n) l <<= 1;
  for (int i = 0; i < la; i++) tmpa[i] = A[i];
  for (int i = la; i < l; i++) tmpa[i] = 0;
  for (int i = 0; i < lb; i++) tmpb[i] = B[i];
  for (int i = lb; i < l; i++) tmpb[i] = 0;
  NTT(tmpa, l), NTT(tmpb, l);
  for (int i = 0; i < l; i++) tmpa[i] = 1ll * tmpa[i] * tmpb[i] % mod;
  reverse(tmpa + 1, tmpa + l);
  NTT(tmpa, l);
  int iv = mod - (mod - 1) / l;
  if (K + 1 < n) n = K + 1;
  c.resize(n);
  for (int i = 0; i < n; i++) c[i] = 1ll * tmpa[i] * iv % mod;
}
void solve(int rot, int lt, int rt) {
  if (lt == rt) {
    a[rot].resize(2);
    a[rot][0] = 1, a[rot][1] = b[lt];
    return;
  }
  int mid = (lt + rt) >> 1;
  solve(rot << 1, lt, mid), solve(rot << 1 | 1, mid + 1, rt);
  mul(a[rot << 1], a[rot << 1 | 1], a[rot]);
}
int main() {
  scanf("%d%d", &n, &K);
  prework(n);
  for (int i = 1; i <= n; i++) scanf("%d", tw + i);
  scanf("%d", &qn);
  for (int i = 1, opt, l, r, q, d; i <= qn; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d%d", &q, &l, &d);
      for (int j = 1; j <= n; j++) b[j] = q - tw[j];
      b[l] = q - d;
      solve(1, 1, n);
      printf("%d\n", (a[1][K] + mod) % mod);
    } else {
      scanf("%d%d%d%d", &q, &l, &r, &d);
      for (int j = 1; j <= n; j++) b[j] = q - tw[j];
      for (int j = l; j <= r; j++) b[j] -= d;
      solve(1, 1, n);
      printf("%d\n", (a[1][K] + mod) % mod);
    }
  }
}
