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;
template <class T, class T1>
int chkmin(T &x, const T1 &y) {
  return x > y ? x = y, 1 : 0;
}
template <class T, class T1>
int chkmax(T &x, const T1 &y) {
  return x < y ? x = y, 1 : 0;
}
const int MAXN = (1 << 15);
const int mod = 998244353;
int n, k;
int _a[MAXN], a[MAXN];
void read() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> _a[i];
  }
}
int perm[MAXN], iperm[MAXN];
int pw(int x, int p) {
  int r = 1;
  while (p) {
    if (p & 1) r = r * 1ll * x % mod;
    x = x * 1ll * x % mod;
    p >>= 1;
  }
  return r;
}
void precompute() {
  perm[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    perm[i] = i * 1ll * perm[i - 1] % mod;
  }
  iperm[MAXN - 1] = pw(iperm[MAXN - 1], mod - 2);
  for (int i = MAXN - 2; i >= 0; i--) {
    iperm[i] = iperm[i + 1] * 1ll * (i + 1) % mod;
  }
}
int dp[MAXN];
int solve_fast() {
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = min(i + 1, k); j >= max(1, k - (n - i - 1)); --j) {
      dp[j] = (dp[j] + dp[j - 1] * 1ll * a[i]) % mod;
    }
  }
  return dp[k];
}
void solve() {
  precompute();
  int q;
  cin >> q;
  while (q--) {
    int type, x;
    cin >> type >> x;
    for (int i = 0; i < n; i++) a[i] = _a[i];
    if (type == 1) {
      int i, d;
      cin >> i >> d;
      a[i - 1] = d;
    } else {
      int l, r, d;
      cin >> l >> r >> d;
      for (int i = l - 1; i < r; i++) a[i] += d, a[i] %= mod;
    }
    for (int i = 0; i < n; i++) {
      a[i] = (x - a[i] + mod) % mod;
    }
    cout << solve_fast() << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}