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 mod = 998244353, g = 3, ig = 332748118;
long long fpow(long long a, long long b, long long c) {
  long long ret = 1 % c;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
void ntt(vector<long long> &v, vector<int> &rev, int opt) {
  int tot = v.size();
  for (int i = 0; i < tot; i++)
    if (i < rev[i]) swap(v[i], v[rev[i]]);
  for (int mid = 1; mid < tot; mid <<= 1) {
    long long wn = fpow(opt == 1 ? g : ig, (mod - 1) / (mid << 1), mod);
    for (int j = 0; j < tot; j += mid << 1) {
      long long w = 1;
      for (int k = 0; k < mid; k++) {
        long long x = v[j + k], y = v[j + mid + k] * w % mod;
        v[j + k] = (x + y) % mod, v[j + mid + k] = (x - y + mod) % mod;
        w = w * wn % mod;
      }
    }
  }
  if (opt == -1) {
    long long itot = fpow(tot, mod - 2, mod);
    for (int i = 0; i < tot; i++) v[i] = v[i] * itot % mod;
  }
}
vector<long long> convolution(
    vector<long long> &a, int cnta, vector<long long> &b, int cntb,
    const function<long long(long long, long long)> &calc) {
  int bit = 0, tot = 1;
  while (tot <= 2 * max(cnta, cntb)) bit++, tot <<= 1;
  vector<int> rev(tot);
  for (int i = 0; i < tot; i++)
    rev[i] = rev[i >> 1] >> 1 | (i & 1) << (bit - 1);
  vector<long long> foo(tot), bar(tot);
  for (int i = 0; i < cnta; i++) foo[i] = a[i];
  for (int i = 0; i < cntb; i++) bar[i] = b[i];
  ntt(foo, rev, 1), ntt(bar, rev, 1);
  for (int i = 0; i < tot; i++) foo[i] = calc(foo[i], bar[i]);
  ntt(foo, rev, -1);
  return foo;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, K;
  cin >> n >> K;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int m;
  cin >> m;
  while (m--) {
    int opt;
    cin >> opt;
    vector<long long> b = a;
    if (opt == 1) {
      int q, x, y;
      cin >> q >> x >> y;
      x--;
      b[x] = y;
      for (int i = 0; i < n; i++) {
        b[i] = (q - b[i] + mod) % mod;
      }
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      l--, r--;
      for (int i = l; i <= r; i++) {
        b[i] = (b[i] + d) % mod;
      }
      for (int i = 0; i < n; i++) {
        b[i] = (q - b[i] + mod) % mod;
      }
    }
    function<vector<long long>(int, int)> solve = [&](int l, int r) {
      if (l == r) {
        return vector<long long>{1, b[l]};
      }
      int mid = l + r >> 1;
      vector<long long> ls = solve(l, mid);
      vector<long long> rs = solve(mid + 1, r);
      return convolution(ls, mid - l + 2, rs, r - mid + 1,
                         [&](long long a, long long b) { return a * b % mod; });
    };
    vector<long long> ans = solve(0, n - 1);
    cout << ans[K] << endl;
  }
  return 0;
}
