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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
  return os << "(" << x.first << "," << x.second << ")";
}
const int MOD = 998244353;
long long modex(long long a, long long p, long long m = MOD) {
  if (!p) return 1;
  if (p & 1) return a * modex(a, p - 1, m) % m;
  long long v = modex(a, p >> 1, m);
  return v * v % m;
}
namespace fft {
const int mod = 998244353;
const int gen = 3;
vector<int> rev(1, 0), roots;
int base = 0;
void precompute(int nbase) {
  if (nbase <= base) return;
  rev.resize(1 << nbase);
  for (int i = 0; i < (1 << nbase); ++i) {
    rev[i] = rev[i >> 1] >> 1 | ((i & 1) << nbase - 1);
  }
  roots.resize(1 << nbase);
  for (; base < nbase; ++base) {
    int len = 1 << base;
    long long root = modex(gen, (mod - 1) >> base + 1, mod);
    roots[len] = 1;
    for (int i = 1; i < len; ++i)
      roots[len + i] = roots[len + i - 1] * root % mod;
  }
}
void fft(vector<int>& a) {
  int n = a.size();
  assert((n & (n - 1)) == 0);
  int zeros = __builtin_ctz(n);
  precompute(zeros);
  int shift = base - zeros;
  for (int i = 0; i < n; ++i) {
    if (i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]);
  }
  for (int len = 1; len < n; len <<= 1)
    for (int i = 0; i < n; i += 2 * len)
      for (int j = 0; j < len; ++j) {
        int u = a[i + j], v = (long long)a[i + j + len] * roots[len + j] % mod;
        a[i + j] = u + v >= mod ? u + v - mod : u + v;
        a[i + j + len] = u - v < 0 ? u - v + mod : u - v;
      }
}
vector<int> multiply(vector<int> a, vector<int> b) {
  if (a.empty() || b.empty()) return {};
  int sz = a.size() + b.size() - 1;
  int n = sz == 1 ? 1 : 1 << (32 - __builtin_clz(sz - 1));
  bool eq = (a == b);
  a.resize(n);
  fft(a);
  if (eq)
    b = a;
  else {
    b.resize(n);
    fft(b);
  }
  long long inv = modex(n, mod - 2, mod);
  for (int i = 0; i < n; ++i) a[i] = (long long)a[i] * b[i] % mod * inv % mod;
  fft(a);
  reverse(a.begin() + 1, a.end());
  a.resize(sz);
  return a;
}
vector<int> multiply(const vector<int>& a, const vector<int>& b, int m) {
  assert(m == mod);
  return multiply(a, b);
}
};  // namespace fft
vector<int> B;
vector<int> multiplyAll(int s, int e) {
  if (s == e) {
    return {1, B[s]};
  }
  int m = (s + e) >> 1;
  auto a = multiplyAll(s, m);
  auto b = multiplyAll(m + 1, e);
  return fft::multiply(a, b);
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) x += MOD;
  return x;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> A(n);
  B = vector<int>(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
    A[i] %= MOD;
  }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int t, q, a, b, d;
    scanf("%d%d%d%d", &t, &q, &a, &b);
    if (t == 1) {
      b %= MOD;
      for (int i = 0; i < n; ++i) B[i] = sub(q, A[i]);
      B[a - 1] = sub(q, b);
    } else {
      scanf("%d", &d);
      d %= MOD;
      for (int i = 0; i < n; ++i) B[i] = sub(q, A[i]);
      for (int i = a - 1; i < b; ++i) B[i] = sub(B[i], d);
    }
    auto p = multiplyAll(0, n - 1);
    printf("%d\n", p[k]);
  }
}