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>
const int N = 2e4 + 5;
const long long mod = 998244353, ot = 3;
long long qpow(long long x, long long y) {
  long long res = 1;
  x %= mod;
  y %= (mod - 1);
  if (y < 0) y += mod - 1;
  for (; y; x = x * x % mod, y >>= 1)
    if (y & 1) res = res * x % mod;
  return res;
}
void NTT(std::vector<long long> &a, int n, bool inv = 0) {
  for (int i = 0, j = 0; i < n; i++) {
    if (j > i) std::swap(a[i], a[j]);
    int k = n;
    while (j & (k >>= 1)) j &= ~k;
    j |= k;
  }
  long long p = inv ? -(mod - 1) / 2 : (mod - 1) / 2;
  for (int i = 1; i < n; i <<= 1) {
    long long c = qpow(ot, p / i), o = 1;
    for (int k = 0; k < i; k++, o = o * c % mod) {
      for (int j = k; j < n; j += i << 1) {
        int e = j + i;
        long long t = o * a[e] % mod;
        a[e] = (a[j] - t + mod) % mod;
        a[j] = (a[j] + t) % mod;
      }
    }
  }
  if (inv) {
    p = qpow(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = a[i] * p % mod;
  }
}
std::vector<long long> mul(std::vector<long long> a, std::vector<long long> b) {
  int la = a.size(), lb = b.size();
  int n = 1;
  while (n < la + lb - 1) n <<= 1;
  std::vector<long long> ans(n, 0);
  a.resize(n, 0);
  b.resize(n, 0);
  NTT(a, n, false);
  NTT(b, n, false);
  for (int i = 0; i < n; i++) ans[i] = a[i] * b[i] % mod;
  NTT(ans, n, true);
  return ans;
}
long long a[N], b[N];
std::vector<long long> solve(int l, int r) {
  std::vector<long long> ans;
  if (l == r) {
    ans.resize(2, 0);
    ans[0] = 1;
    ans[1] = b[l];
    return ans;
  }
  int mid = (l + r) >> 1;
  ans = mul(solve(l, mid), solve(mid + 1, r));
  while (ans.size() > 1 && ans.back() == 0) ans.pop_back();
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  int T;
  scanf("%d", &T);
  for (int op, l, r, d, q; T--;) {
    scanf("%d%d", &op, &q);
    if (op == 1) {
      scanf("%d%d", &l, &r);
      for (int i = 1; i <= n; i++) {
        if (i != l)
          b[i] = (q - a[i] + mod + mod) % mod;
        else
          b[i] = (q - r + mod + mod) % mod;
      }
    } else {
      scanf("%d%d%d", &l, &r, &d);
      for (int i = 1; i <= n; i++) {
        if (i < l || i > r)
          b[i] = (q - a[i] + mod + mod) % mod;
        else
          b[i] = (q - a[i] - d + mod + mod + mod) % mod;
      }
    }
    std::vector<long long> ans = solve(1, n);
    printf("%lld\n", ans.size() > k ? ans[k] : 0LL);
  }
  return 0;
}
