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;
long long modx(long long Base, long long exponent) {
  long long ans = 1;
  if (Base == 1) return Base;
  while (exponent) {
    if (exponent & 1) ans = (ans * Base) % 998244353LL;
    Base = (Base * Base) % 998244353LL;
    exponent = exponent >> 1;
  }
  return ans;
}
long long inmodx(long long num) { return (modx(num, 998244353LL - 2LL)); }
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.second < b.second) return 1;
  if (a.second > b.second) return 0;
  if (a.first <= b.first) return 1;
  return 0;
}
const int N = (1e5) + 9;
const int M = (N << 2) + 9;
const int LOGN = ((int)log2(N)) + 3;
const int LOGM = ((int)log2(M)) + 3;
const int BUCK = 900;
const int primitive_root = 3;
const int root_pw = 1 << 20;
const int root = modx(primitive_root, (998244353LL - 1) / root_pw);
const int root_1 = inmodx(root);
int mul(const int &a, const int &b) {
  long long ans = 1LL * a * 1LL * b;
  if (ans >= 998244353LL) return ans % 998244353LL;
  return ans;
}
void fft(vector<int> &a, bool invert) {
  int n = (int)a.size();
  for (int i = 1, j = 0; i < n; ++i) {
    int bit = n >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = int(wlen * 1LL * wlen % 998244353LL);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; ++j) {
        int u = a[i + j], v = int(a[i + j + len / 2] * 1LL * w % 998244353LL);
        a[i + j] = u + v < 998244353LL ? u + v : u + v - 998244353LL;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + 998244353LL;
        w = int(w * 1LL * wlen % 998244353LL);
      }
    }
  }
  if (invert) {
    int nrev = inmodx(n);
    for (int i = 0; i < n; ++i) a[i] = int(a[i] * 1LL * nrev % 998244353LL);
  }
}
void multiply(vector<int> &a, vector<int> b) {
  int n = a.size() + b.size() - 1, i;
  int m = 1;
  while (m < n) m <<= 1;
  a.resize(m);
  b.resize(m);
  fft(a, 0);
  fft(b, 0);
  for (i = 0; i < m; ++i) a[i] = mul(a[i], b[i]);
  fft(a, 1);
  a.resize(n);
}
long long a[N], b[N];
vector<int> v[N];
void dfs(int l, int r) {
  if (l == r) {
    v[l] = {1, (int)b[l]};
    return;
  }
  int mid = (l + r) >> 1;
  dfs(l, mid);
  dfs(mid + 1, r);
  multiply(v[l], v[mid + 1]);
  v[mid + 1].clear();
  return;
}
void solve() {
  int n, k, i, Q, t, l, r, ind;
  long long d, ans, q, diff;
  cin >> n >> k;
  for (i = (1); i < (n + 1); i++) {
    cin >> a[i];
    a[i] %= 998244353LL;
  }
  cin >> Q;
  while (Q--) {
    cin >> t;
    if (t == 1) {
      cin >> q >> ind >> d;
      q %= 998244353LL;
      d %= 998244353LL;
      diff = (d - a[ind] + 998244353LL) % 998244353LL;
      a[ind] = (a[ind] + diff) % 998244353LL;
    } else {
      cin >> q >> l >> r >> diff;
      q %= 998244353LL;
      diff %= 998244353LL;
      for (i = (l); i < (r + 1); i++) a[i] = (a[i] + diff) % 998244353LL;
    }
    for (i = (1); i < (n + 1); i++)
      b[i] = (q + 998244353LL - a[i]) % 998244353LL;
    if (t == 1)
      a[ind] = (a[ind] + 998244353LL - diff) % 998244353LL;
    else
      for (i = (l); i < (r + 1); i++)
        a[i] = (a[i] + 998244353LL - diff) % 998244353LL;
    dfs(1, n);
    cout << v[1][k] << '\n';
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  clock_t clk;
  clk = clock();
  int t = 1, cs;
  for (cs = (1); cs < (t + 1); cs++) {
    solve();
  }
  clk = clock() - clk;
  cerr << fixed << setprecision(6) << "Time: " << ((double)clk) / CLOCKS_PER_SEC
       << "\n";
  return 0;
}
