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
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int P = 3;
int pow(int a, int k) {
  if (k == 0) return 1;
  if (k == 1) return a;
  long long res = pow(a, k / 2);
  res = res * res % mod;
  if (k & 1) res = res * a % mod;
  return res;
}
void ntt(vector<int> &v, int inv = 0) {
  int n = v.size();
  for (int i = 1, j = 0; i < n; i++) {
    int s = n;
    do {
      s >>= 1;
      j ^= s;
    } while (~j & s);
    if (i < j) swap(v[i], v[j]);
  }
  for (int i = 2; i <= n; i <<= 1) {
    long long xi = pow(P, (mod - 1) / i);
    if (inv) xi = pow(xi, mod - 2);
    int h = i >> 1;
    for (int j = 0; j < n; j += i) {
      long long w = 1;
      for (int k = 0; k < h; k++) {
        long long p = v[j + k], q = v[j + k + h];
        v[j + k] = (p + q * w) % mod;
        v[j + k + h] = (p - q * w) % mod;
        if (v[j + k + h] < 0) v[j + k + h] += mod;
        w = w * xi % mod;
      }
    }
  }
  if (inv) {
    long long iv = pow(n, mod - 2);
    for (int i = 0; i < n; i++) {
      v[i] = v[i] * iv % mod;
    }
  }
}
int a[20010], b[20010];
vector<int> po[16][20010];
vector<int> mul(vector<int> &v, vector<int> &w) {
  vector<int> r(v.size() + w.size() - 1);
  for (int i = 0; i < v.size(); i++) {
    for (int j = 0; j < w.size(); j++) {
      r[i + j] = (r[i + j] + (long long)v[i] * w[j]) % mod;
    }
  }
  return r;
}
int main() {
  int n, k, q;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  scanf("%d", &q);
  while (q--) {
    int ty, e, l, r, x;
    scanf("%d%d%d", &ty, &e, &l);
    --l;
    for (int i = 0; i < n; i++) {
      b[i] = a[i];
    }
    if (ty == 1) {
      scanf("%d", &x);
      b[l] = x;
    } else {
      scanf("%d%d", &r, &x);
      for (int i = l; i < r; i++) {
        b[i] += x;
      }
    }
    for (int i = 0; i < n; i++) {
      b[i] = e - b[i];
      b[i] = (b[i] % mod + mod) % mod;
      po[0][i] = vector<int>(2, 1);
      po[0][i][0] = b[i];
    }
    int s = 0;
    for (int i = 0; n >> i; i++) {
      int h = 1 << i;
      for (int j = 0; j < n; j += h * 2) {
        if (j + h >= n) {
          po[i + 1][j] = po[i][j];
          po[i + 1][j].resize(h * 4);
          continue;
        }
        po[i][j].resize(h * 4);
        po[i][j + h].resize(h * 4);
        ntt(po[i][j]);
        ntt(po[i][j + h]);
        po[i + 1][j] = vector<int>(h * 4);
        for (int k = 0; k < h * 4; k++) {
          po[i + 1][j][k] = (long long)po[i][j][k] * po[i][j + h][k] % mod;
        }
        ntt(po[i + 1][j], 1);
      }
      s++;
    }
    printf("%d\n", po[s][0][n - k]);
  }
  return 0;
}
