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;
const long long mod = 998244353, mod2 = 1e9 + 9, p1 = 5807, p2 = 8573;
const long long N = 2e4 + 5, M = 21;
const long double pi = acos(-1);
long long power(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void MOD(long long* x, long long mod) {
  if (*x >= mod) {
    *x -= (*x / mod) * mod;
  }
}
const long long root = 15311432;
const long long root_1 = 469870224;
const long long root_pw = 1 << 23;
void fft(vector<long long>& a, bool invert) {
  long long n = a.size();
  for (long long i = 1, j = 0; i < n; i++) {
    long long bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (long long len = 2; len <= n; len <<= 1) {
    long long wlen = invert ? root_1 : root;
    for (long long i = len; i < root_pw; i <<= 1)
      wlen = (long long)((1LL * wlen * wlen) % mod);
    for (long long i = 0; i < n; i += len) {
      long long w = 1;
      for (long long j = 0; j < len / 2; j++) {
        long long u = a[i + j],
                  v = (long long)((a[i + j + len / 2] * 1LL * w) % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = (long long)((w * 1LL * wlen) % mod);
      }
    }
  }
  if (invert) {
    long long n_1 = power(n, mod - 2, mod);
    for (long long& x : a) x = (long long)(x * 1LL * n_1 % mod);
  }
}
vector<long long> multiply(vector<long long> const& a,
                           vector<long long> const& b) {
  vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  long long n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (long long i = 0; i < n; i++) {
    fa[i] = (long long)((fa[i] * 1LL * fb[i]) % mod);
  }
  fft(fa, true);
  vector<long long> result(n);
  for (long long i = 0; i < n; i++) {
    result[i] = fa[i] % mod;
  }
  return result;
}
void clean(vector<long long>& v) {
  while (v.size() && v[v.size() - 1] == 0) {
    v.pop_back();
  }
}
vector<long long> v, vec;
vector<long long> fun(long long start, long long end) {
  if (start == end) {
    vector<long long> vv = {vec[start], 1};
    return vv;
  } else {
    long long mid = (start + end) / 2;
    vector<long long> v1 = fun(start, mid);
    vector<long long> v2 = fun(mid + 1, end);
    vector<long long> vv = multiply(v1, v2);
    clean(vv);
    return vv;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  v.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long q;
  cin >> q;
  while (q--) {
    long long type;
    cin >> type;
    long long p;
    if (type == 1) {
      long long idx, val;
      cin >> p >> idx >> val;
      idx--;
      vec = v;
      vec[idx] = val;
    } else {
      long long l, r, d;
      cin >> p >> l >> r >> d;
      l--;
      r--;
      vec = v;
      for (long long i = l; i <= r; i++) {
        vec[i] += d;
      }
    }
    for (long long i = 0; i < n; i++) {
      vec[i] = p - vec[i];
    }
    vector<long long> res = fun(0, n - 1);
    cout << (res[n - k] + mod) % mod << "\n";
  }
  return 0;
}