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 A = 119, B = 23, P = A << B | 1, R = 3;
const int SZ = 20, N = 1 << SZ;
int Pow(int x, int y) {
  int r = 1;
  while (y) {
    if (y & 1) r = (long long)r * x % P;
    x = (long long)x * x % P;
    y >>= 1;
  }
  return r;
}
void FFT(vector<long long> &a, bool f) {
  int i, j, k, x, y, z;
  j = 0;
  int SZ = a.size();
  for (i = 1; i < SZ; i++) {
    for (k = SZ >> 1; j >= k; k >>= 1) j -= k;
    j += k;
    if (i < j) {
      k = a[i];
      a[i] = a[j];
      a[j] = k;
    }
  }
  for (i = 1; i < SZ; i <<= 1) {
    x = Pow(f ? Pow(R, P - 2) : R, P / i >> 1);
    for (j = 0; j < SZ; j += i << 1) {
      y = 1;
      for (k = 0; k < i; k++) {
        z = (long long)a[i | j | k] * y % P;
        a[i | j | k] = a[j | k] - z;
        if (a[i | j | k] < 0) a[i | j | k] += P;
        a[j | k] += z;
        if (a[j | k] >= P) a[j | k] -= P;
        y = (long long)y * x % P;
      }
    }
  }
  if (f) {
    j = Pow(SZ, P - 2);
    for (i = 0; i < SZ; i++) a[i] = (long long)a[i] * j % P;
  }
}
vector<long long> multiply(vector<long long> &a, vector<long long> &b) {
  int n = 1;
  while (n < max(((int)(a).size()), ((int)(b).size()))) n <<= 1;
  n <<= 1;
  a.resize(n, 0);
  b.resize(n, 0);
  FFT(a, false);
  FFT(b, false);
  for (int i = 0; i < n; i++) a[i] = (long long)a[i] * b[i] % P;
  FFT(a, true);
  return a;
}
int in[N], in2[N], a[N];
vector<long long> dnc(int l, int r) {
  if (l == r) {
    vector<long long> ret(2);
    ret[0] = 1;
    ret[1] = a[l];
    return ret;
  } else {
    int mid = (l + r) / 2;
    vector<long long> x = dnc(l, mid), y = dnc(mid + 1, r);
    return multiply(x, y);
  }
}
void solve() {
  int n;
  scanf("%d ", &n);
  int k;
  scanf("%d ", &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d ", &in2[i]);
  }
  int q;
  scanf("%d ", &q);
  while (q--) {
    int type, q, l, r;
    scanf("%d %d %d %d ", &type, &q, &l, &r);
    for (int i = 0; i < n; ++i) {
      in[i] = in2[i];
    }
    if (type == 2) {
      int d;
      scanf("%d ", &d);
      for (int i = l - 1; i <= r - 1; ++i) {
        in[i] = in[i] + d;
      }
    } else {
      in[l - 1] = r;
    }
    for (int i = 0; i < n; ++i) {
      a[i] = ((q - in[i]) % P + P) % P;
    }
    vector<long long> ret = dnc(0, n - 1);
    printf("%lld\n", ret[k]);
  }
}
int main() {
  solve();
  return 0;
}