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 N = 2e5 + 10;
const int mod = 998244353;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
int qpow(int a, int p, int mod) {
  int res = 1;
  while (p) {
    if (p & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    p >>= 1;
  }
  return res;
}
int inc(int x, int y, int mo) {
  if (y < 0) y += mo;
  if (x + y >= mo) x -= mo;
  return x + y;
}
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int rev[N];
void NTT(int *A, int n, int inv) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(A[i], A[rev[i]]);
  for (int l = 1; l < n; l <<= 1) {
    int tt = qpow(gi, (P - 1) / (l << 1), P);
    int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
    for (int i = 0; i < n; i += (l << 1)) {
      int omega = 1;
      for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P) {
        int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
        A[i + j] = inc(x, y, P);
        A[i + j + l] = inc(x, -y, P);
      }
    }
  }
  int invv = qpow(n, P - 2, P);
  if (inv == -1)
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * invv % P;
}
int NTTinit(int n, int m) {
  int ML = 1, bit = 0;
  while (ML <= n + m) ML <<= 1, bit++;
  for (int i = 0; i < ML; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  return ML;
}
void NTTX(int *a, int n, int *b, int m) {
  int ML = NTTinit(n, m);
  NTT(a, ML, 1);
  NTT(b, ML, 1);
  for (int i = 0; i < ML; i++) a[i] = 1ll * a[i] * b[i] % P;
  NTT(a, ML, -1);
}
int b[N];
void solve(int l, int r, int *res) {
  if (l == r) {
    res[0] = 1;
    res[1] = b[l];
    return;
  }
  int mid = (l + r) >> 1;
  int n = (mid - l + 1), m = (r - mid + 1);
  int ML = NTTinit(n, m);
  int L[ML], R[ML];
  memset(L, 0, sizeof(L));
  memset(R, 0, sizeof(R));
  solve(l, mid, L);
  solve(mid + 1, r, R);
  NTTX(L, n, R, m);
  for (int i = 0; i <= r - l + 1; i++) res[i] = L[i];
}
int a[N], C[N];
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int Q = read();
  while (Q--) {
    int op = read();
    memset(C, 0, sizeof(C));
    if (op == 1) {
      int q = read(), ii = read(), d = read();
      for (int i = 1; i <= n; i++)
        b[i] = (q - (i == ii ? d : a[i]) + mod) % mod;
    } else {
      int q = read(), l = read(), r = read(), d = read();
      for (int i = 1; i <= n; i++)
        b[i] = (q - ((i <= r && i >= l) ? (d + a[i]) : a[i]) + mod) % mod;
    }
    solve(1, n, C);
    printf("%d\n", C[k]);
  }
}
