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;
using ll = long long;
using Vi = vector<int>;
using Pii = pair<int, int>;
template <class t, class u, class... w>
auto operator<<(t& o, u a) -> decltype(a.print(), o) {
  o << "{";
  a.print();
  o << "}";
  return o;
};
template <class t, class u, class... w>
auto operator<<(t& o, u a) -> decltype(a.second, o) {
  o << "(" << a.first << ", " << a.second << ")";
  return o;
};
template <class t, class u, class... w>
auto operator<<(t& o, u a) -> decltype((a).begin(), (a).end(), o) {
  o << "[";
  auto d = "";
  for (auto i : a) (o << d << i, d = ", ");
  o << "]";
  return o;
};
void DD(...) {}
template <class t, class u, class... w>
void DD(t s, u a, w... k) {
  int b = 44;
  while (*s && *s != b) {
    b += (*s == 40 ? 50 : *s == 41 ? -50 : 0);
    cerr << *s++;
  }
  cerr << ": " << a << *s++;
  DD(s, k...);
}
template <class t, class u, class... w>
vector<t> span(const t* a, u n) {
  return {a, a + n};
}
constexpr ll MOD = 998244353;
constexpr ll ROOT = 31;
Vi tmp;
ll modInv(ll a, ll m) {
  if (a == 1) return 1;
  return ((a - modInv(m % a, a)) * m + 1) / a;
}
template <int dir>
void fft(vector<ll>& buf) {
  assert(__builtin_popcount(int((buf).size())) == 1);
  int n = int((buf).size()), bits = 31 - __builtin_clz(n);
  int i = (dir > 0 ? 0 : bits - 1);
  vector<ll> bases = {1};
  ll c = ROOT;
  for (int k = (bits); k < (23); k++) c = c * c % MOD;
  for (int k = (0); k < (n); k++) bases.push_back(bases.back() * c % MOD);
  for (; i >= 0 && i < bits; i += dir) {
    int shift = 1 << (bits - i - 1);
    for (int j = (0); j < (1 << i); j++)
      for (int k = (0); k < (shift); k++) {
        int a = (j << (bits - i)) | k, b = a | shift;
        auto v1 = buf[a], v2 = buf[b];
        auto base = bases[(dir * (k << i)) & (n - 1)];
        if (dir > 0) {
          buf[b] = (v1 - v2) * base % MOD;
        } else {
          v2 *= base;
          buf[b] = (v1 - v2) % MOD;
        }
        buf[a] = (v1 + v2) % MOD;
      }
  }
  if (dir < 0) {
    ll second = modInv(n, MOD);
    for (auto& first : (buf)) first = first * second % MOD;
  }
  for (auto& first : (buf))
    if (first < 0) first += MOD;
}
void convolve(vector<ll>& a, vector<ll> b) {
  int len = int((a).size()) + int((b).size()) - 1;
  int n = 1 << (32 - __builtin_clz(len));
  a.resize(n);
  b.resize(n);
  fft<1>(a);
  fft<1>(b);
  for (int i = (0); i < (n); i++) a[i] = a[i] * b[i] % MOD;
  fft<-1>(a);
  a.resize(len);
}
vector<ll> solve(int b, int e) {
  if (b >= e) return {1};
  if (b + 1 == e) return {1, tmp[b] % MOD};
  int m = (b + e) / 2;
  auto vec1 = solve(b, m);
  auto vec2 = solve(m, e);
  convolve(vec1, vec2);
  return vec1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(18);
  int n, k;
  cin >> n >> k;
  Vi elems(n);
  for (auto& e : (elems)) cin >> e;
  int queries;
  cin >> queries;
  tmp.resize(n);
  while (queries--) {
    int t;
    cin >> t;
    if (t == 1) {
      int q, i, d;
      cin >> q >> i >> d;
      for (int j = (0); j < (n); j++) {
        tmp[j] = q - (i == j + 1 ? d : elems[j]);
      }
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      for (int j = (0); j < (n); j++) {
        tmp[j] = q - elems[j] - (j + 1 >= l && j + 1 <= r ? d : 0);
      }
    }
    auto poly = solve(0, n);
    ll ans = poly[k];
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
  }
}