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 long long mod = 998244353;
const int maxn = 4e4 + 9;
const int inf = 0x3f3f3f3f;
long long rd() {
  long long ans = 0;
  char last = ' ', ch = getchar();
  while (!(ch >= '0' && ch <= '9')) last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
void rader(vector<long long>& a, int n) {
  for (int i = 1, j = n >> 1; i < n - 1; i++) {
    if (i < j) a[i] ^= a[j], a[j] ^= a[i], a[i] ^= a[j];
    int k = n >> 1;
    for (; j >= k; k >>= 1) j -= k;
    if (j < k) j += k;
  }
}
const int G = 3;
void NTT(vector<long long>& a, int n, int on) {
  rader(a, n);
  for (int h = 2; h <= n; h <<= 1) {
    int hh = h >> 1,
        wn = Pow(G, on == -1 ? mod - 1 - (mod - 1) / h : (mod - 1) / h);
    for (int i = 0; i < n; i += h) {
      long long w = 1;
      for (int j = i; j < i + hh; j++) {
        int x = a[j], y = w * a[j + hh] % mod;
        a[j] = (x + y) % mod;
        a[j + hh] = (x - y + mod) % mod;
        w = w * wn % mod;
      }
    }
  }
  if (on == -1) {
    int inv = Pow(n, mod - 2);
    for (int i = (int)(0); i <= (int)(n - 1); i++) a[i] = a[i] * inv % mod;
  }
}
int n, k;
long long a[maxn], b[maxn];
vector<long long> divide(int l, int r) {
  if (l == r) {
    vector<long long> res(2);
    res[0] = 1;
    res[1] = b[l];
    return res;
  }
  int mid = l + r >> 1;
  auto L = divide(l, mid), R = divide(mid + 1, r);
  int len = 1;
  while (len < 2 * (r - l + 1)) len <<= 1;
  while (L.size() < len) L.push_back(0);
  while (R.size() < len) R.push_back(0);
  NTT(L, len, 1);
  NTT(R, len, 1);
  for (int i = (int)(0); i <= (int)(len - 1); i++) {
    L[i] = L[i] * R[i] % mod;
  }
  NTT(L, len, -1);
  return L;
}
int main() {
  n = rd(), k = rd();
  for (int i = (int)(1); i <= (int)(n); i++) a[i] = rd();
  int _ = rd();
  while (_--) {
    int op = rd();
    long long q = rd();
    if (op == 1) {
      int I = rd();
      long long d = rd();
      for (int i = (int)(1); i <= (int)(n); i++) {
        if (i == I) {
          b[i] = (q - d + mod) % mod;
        } else {
          b[i] = (q - a[i] + mod) % mod;
        }
      }
    } else {
      int l = rd(), r = rd();
      long long d = rd();
      for (int i = (int)(1); i <= (int)(n); i++) {
        if (i >= l && i <= r) {
          b[i] = (q - (a[i] + d) + mod) % mod;
        } else {
          b[i] = (q - a[i] + mod) % mod;
        }
      }
    }
    auto ans = divide(1, n);
    printf("%lld\n", (ans[k] + mod) % mod);
  }
  return 0;
}