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;
const int inv2 = 499122177;
int ksm(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % mod;
    y >>= 1;
    x = 1ll * x * x % mod;
  }
  return ret;
}
int wl, wn[33333], inv[33333], rev[33333];
int A[33333];
vector<int> f[20005];
void shai() {
  wl = 1 << 15;
  wn[0] = 1;
  wn[1] = ksm(3, (mod - 1) / wl);
  for (int i = 2; i <= wl; i++) wn[i] = 1ll * wn[i - 1] * wn[1] % mod;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= wl; i <<= 1) inv[i] = 1ll * inv[i >> 1] * inv2 % mod;
}
void NTT(vector<int> &a, int len, int flg) {
  for (int i = 1; i < len; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1, x = wl >> 1; i < len; i <<= 1, x >>= 1)
    for (int j = 0; j < len; j += (i << 1))
      for (int k = j, y = 0; k < i + j; k++, y += x) {
        int tmp = 1ll * a[k + i] * wn[flg == 1 ? y : wl - y] % mod;
        a[k + i] = (a[k] + mod - tmp) % mod;
        a[k] = (a[k] + tmp) % mod;
      }
  if (flg == -1)
    for (int i = 0; i < len; i++) a[i] = 1ll * a[i] * inv[len] % mod;
}
vector<int> mul(vector<int> a, vector<int> b) {
  int ret = a.size() + b.size() - 1, len = 1;
  while (len < ret) len <<= 1;
  a.resize(len);
  b.resize(len);
  for (int i = 1; i < len; i++)
    rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? len >> 1 : 0);
  NTT(a, len, 1);
  NTT(b, len, 1);
  for (int i = 0; i < len; i++) a[i] = 1ll * a[i] * b[i] % mod;
  NTT(a, len, -1);
  a.resize(ret);
  return a;
}
vector<int> solve(int l, int r) {
  if (l == r) return f[l];
  int mid = (l + r) >> 1;
  return mul(solve(l, mid), solve(mid + 1, r));
}
int main() {
  int n, m, Q, i, op, q, l, r, d, x;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    f[i].resize(2);
  }
  shai();
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &op, &q);
    if (op == 1) {
      scanf("%d%d", &x, &d);
      int tmp = A[x];
      A[x] = d;
      for (i = 1; i <= n; i++) {
        f[i][0] = 1;
        f[i][1] = (q + mod - A[i]) % mod;
      }
      A[x] = tmp;
    } else {
      scanf("%d%d%d", &l, &r, &d);
      for (i = 1; i <= n; i++) {
        if (l <= i && i <= r) A[i] = (A[i] + d) % mod;
        f[i][0] = 1;
        f[i][1] = (q + mod - A[i]) % mod;
        if (l <= i && i <= r) A[i] = (A[i] + (mod - d % mod)) % mod;
      }
    }
    vector<int> ans = solve(1, n);
    printf("%d\n", ans[m]);
  }
}
