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 int N = 2e4 + 5;
int aa[N], bb[N];
const int mod = 998244353;
int powr(int a, int b) {
  a %= mod;
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = ((long long)ans * a) % mod;
    }
    a = ((long long)a * a) % mod;
    b >>= 1;
  }
  return ans;
}
int inv(int a, int b) { return powr(a, b - 2); }
class NTT {
  long long p, g, n, logn;

 public:
  NTT(long long P, long long G, long long LOGN) {
    p = P;
    long long g1 = G;
    logn = LOGN;
    n = (1 << LOGN);
    assert((p - 1) % n == 0);
    g = powr(g1, (p - 1) >> logn);
  }
  int bitrevcopy(int i) {
    int ret = 0;
    for (int j = 0; j < logn; j++)
      if (i & (1 << j)) ret |= (1 << (logn - 1 - j));
    return ret;
  }
  void fft(vector<int> &a, bool invert) {
    for (int i = 0; i < n; i++) {
      int br = bitrevcopy(i);
      if (i < br) swap(a[i], a[br]);
    }
    for (int i = 1; i <= logn; i++) {
      long long mul = powr(g, 1 << (logn - i));
      if (invert) mul = inv(mul, p);
      int add = (1 << i);
      for (int k = 0; k < n; k += add) {
        long long w = 1;
        long long lim = (add >> 1);
        for (int j = 0; j < lim; j++) {
          long long u = (w * a[k + j + lim]) % p, v = a[k + j];
          a[k + j] = u + v;
          if (a[k + j] >= p) a[k + j] -= p;
          a[k + j + lim] = v - u;
          if (a[k + j + lim] < 0) a[k + j + lim] += p;
          w = (w * mul) % p;
        }
      }
    }
    if (invert) {
      long long _inv = inv(n, p);
      for (int i = 0; i < n; i++) {
        a[i] = (a[i] * _inv) % p;
      }
    }
  }
};
vector<int> mul(vector<int> &a, vector<int> &b) {
  int sz = 1;
  int LG = 0;
  while (sz < (int)a.size() + (int)b.size()) sz <<= 1, LG++;
  NTT ntt(mod, 3, LG);
  a.resize(sz, 0);
  b.resize(sz, 0);
  ntt.fft(a, 0);
  ntt.fft(b, 0);
  for (int i = 0; i <= sz - 1; i++) {
    a[i] = ((long long)a[i] * b[i]) % mod;
  }
  ntt.fft(a, 1);
  return a;
}
vector<int> query(int l, int r) {
  if (l == r) {
    return {1, bb[l]};
  }
  int m = (l + r) >> 1;
  auto v1 = query(l, m);
  auto v2 = query(m + 1, r);
  return mul(v1, v2);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> aa[i];
    aa[i] %= mod;
  }
  int qq;
  cin >> qq;
  for (int tt = 1; tt <= qq; tt++) {
    int t;
    cin >> t;
    if (t == 1) {
      int q, idx, d;
      cin >> q >> idx >> d;
      for (int i = 1; i <= n; i++) {
        if (i == idx) {
          bb[i] = (q - d + mod);
        } else {
          bb[i] = (q - aa[i] + mod);
        }
        if (bb[i] >= mod) bb[i] -= mod;
      }
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      for (int i = 1; i <= n; i++) {
        if (l <= i && i <= r) {
          bb[i] = (q - aa[i] - d + mod) % mod;
        } else {
          bb[i] = q - aa[i] + mod;
        }
        if (bb[i] >= mod) bb[i] -= mod;
        if (bb[i] < 0) bb[i] += mod;
      }
    }
    auto v = query(1, n);
    cout << v[k] << '\n';
  }
  return 0;
}