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 maxn = 8e4 + 10;
const long long mod = 998244353, modg = 3, modi = 332748118;
int a[maxn], b[maxn];
int pow_mod(int a, int b) {
  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 add(int a, int b) {
  a += b;
  if (a > mod) return a - mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) return a + mod;
  return a;
}
const int maxl = 2e4 + 10;
class nubmer {
 public:
  int rev[maxl << 2], len, pw;
  long long wt[2][maxl << 2];
  void init_0() {
    int len = 1;
    while (len <= maxl) len <<= 1;
    for (int mid = 1; mid < len; mid <<= 1) {
      long long wn1 = pow_mod(modg, (mod - 1) / (mid << 1));
      long long wn2 = pow_mod(modi, (mod - 1) / (mid << 1));
      wt[0][mid] = wt[1][mid] = 1;
      long long wt1 = wn1, wt2 = wn2;
      for (int j = 1; j <= mid; ++j) {
        wt[0][mid + j] = wt1;
        wt1 = wt1 * wn1 % mod;
        wt[1][mid + j] = wt2;
        wt2 = wt2 * wn2 % mod;
      }
    }
  }
  void transform(vector<int> &a, int flag) {
    long long *f = flag == 1 ? wt[0] : wt[1];
    if (a.size() != len) a.resize(len);
    for (int i = 0; i <= len - 1; ++i)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
      for (int r = mid << 1, j = 0; j < len; j += r) {
        long long *p = f + mid;
        for (int k = 0; k < mid; ++k, ++p) {
          int x = a[j + k], y = (*p) * a[j + k + mid] % mod;
          a[j + k + mid] = sub(a[j + k], y), a[j + k] += y;
          if (a[j + k] > mod) a[j + k] -= mod;
        }
      }
    }
    if (flag == -1) {
      long long inv = pow_mod(len, mod - 2);
      for (int i = 0; i <= len - 1; ++i) {
        a[i] = a[i] * inv % mod;
        if (a[i] < 0) a[i] += mod;
      }
    }
  }
  void fix(vector<int> &a) {
    while (!a.empty() && !a.back()) a.pop_back();
  }
  vector<int> mul(vector<int> &a, vector<int> &b) {
    int la = a.size(), lb = b.size();
    if (la + lb <= 100) {
      vector<int> c(la + lb, 0);
      for (int i = 0; i <= la - 1; ++i)
        for (int j = 0; j <= lb - 1; ++j)
          c[i + j] = (c[i + j] + (long long)a[i] * b[j]) % mod;
      return fix(c), c;
    }
    int n = la + lb;
    len = 1, pw = 0;
    while (len <= n) len <<= 1, pw++;
    for (int i = 0; i <= len - 1; ++i)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (pw - 1));
    transform(a, 1);
    transform(b, 1);
    vector<int> c(len, 0);
    for (int i = 0; i <= len - 1; ++i) c[i] = (long long)a[i] * b[i] % mod;
    transform(c, -1);
    return fix(c), c;
  }
} ntt;
int n, k;
int s1[maxn], s2[maxn];
vector<int> solve(int l, int r) {
  if (l == r) return vector<int>{1, s2[l]};
  int mid = (l + r) >> 1;
  vector<int> x = solve(l, mid), y = solve(mid + 1, r);
  return ntt.mul(x, y);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ntt.init_0();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> s1[i];
  }
  int T;
  cin >> T;
  vector<int> tmp;
  while (T--) {
    int flag;
    cin >> flag;
    if (flag == 1) {
      int q, i, d;
      cin >> q >> i >> d;
      for (int j = 1; j <= n; j++) s2[j] = sub(q, s1[j]);
      s2[i] = sub(q, d);
      tmp = solve(1, n);
      if (k >= tmp.size())
        cout << "0\n";
      else
        cout << tmp[k] << "\n";
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      for (int j = 1; j <= n; j++) {
        if (j >= l && j <= r)
          s2[j] = sub(sub(q, s1[j]), d);
        else
          s2[j] = sub(q, s1[j]);
      }
      tmp = solve(1, n);
      if (k >= tmp.size())
        cout << "0\n";
      else
        cout << tmp[k] << "\n";
    }
  }
  return 0;
}