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;
const long long power(long long base, long long expo) {
  long long ans = 1;
  while (expo) {
    if (expo & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    expo >>= 1;
  }
  return ans;
}
class DFT {
 public:
  static const long long mod = 998244353;
  static const long long root = 15311432;
  static const long long root_1 = 469870224;
  static const long long root_pw = 1 << 23;
  static long long reverse(long long base) {
    long long ans = 1, expo = mod - 2;
    while (expo) {
      if (expo % 2) ans = (ans * base) % mod;
      base = (base * base) % mod;
      expo /= 2;
    }
    return ans;
  }
  static void fft(vector<long long> &a, bool invert) {
    long long n = (long long)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)(wlen * 1ll * 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 nrev = reverse(n);
      for (long long i = 0; i < n; ++i)
        a[i] = (long long)(a[i] * 1ll * nrev % mod);
    }
  }
  static void multiply(const vector<long long> &a, const vector<long long> &b,
                       vector<long long> &res) {
    vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
    size_t n = 1;
    while (n < max(a.size(), b.size())) n <<= 1;
    n <<= 1;
    fa.resize(n), fb.resize(n);
    fft(fa, false), fft(fb, false);
    for (size_t i = 0; i < n; ++i)
      fa[i] = (long long)((1ll * fa[i] * fb[i]) % mod);
    fft(fa, true);
    res.resize(n);
    for (size_t i = 0; i < n; ++i) res[i] = fa[i];
  }
};
long long n, k, m, ch, q, l, r, d;
vector<long long> v;
void multiply(long long st, long long en, vector<long long> &l) {
  if (st == en) {
    l = {1, ((q - v[st]) % mod + mod) % mod};
  } else {
    long long mid = (st + en) >> 1;
    vector<long long> r;
    multiply(st, mid, l);
    multiply(mid + 1, en, r);
    DFT::multiply(l, r, l);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  v.resize(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  cin >> m;
  while (m--) {
    cin >> ch >> q >> l >> r;
    vector<long long> dp;
    if (ch == 1) {
      l--;
      swap(v[l], r);
      multiply(0, n - 1, dp);
      swap(v[l], r);
    } else {
      cin >> d;
      r--;
      l--;
      for (long long i = l; i <= r; i++) {
        if (i >= l && i <= r) v[i] += d;
      }
      multiply(0, n - 1, dp);
      for (long long i = l; i <= r; i++) {
        if (i >= l && i <= r) v[i] -= d;
      }
    }
    cout << dp[k] << '\n';
  }
  return 0;
}