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;
namespace ntt {
const int max_lev = 15;
const int max_n = 1 << max_lev, mod = 998244353;
int mul(int x, int y) { return (1LL * x * y) % mod; }
int power(int x, int y) {
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return power(mul(x, x), y / 2);
  }
  return mul(x, power(x, y - 1));
}
int find_root() {
  for (int root = 2;; ++root) {
    if (power(root, max_n) == 1) {
      int cur = 1;
      bool ok = true;
      for (int i = 1; i < max_n; ++i) {
        cur = mul(cur, root);
        if (cur == 1) {
          ok = false;
          break;
        }
      }
      if (ok) {
        return root;
      }
    }
  }
}
int pw[max_n], rpw[max_n], rev[max_n];
void init() {
  int root = find_root();
  int rroot = power(root, mod - 2);
  pw[0] = rpw[0] = 1;
  for (int i = 1; i < max_n; ++i) {
    pw[i] = mul(pw[i - 1], root);
    rpw[i] = mul(rpw[i - 1], rroot);
    rev[i] = (rev[i / 2] / 2) + ((i & 1) << (max_lev - 1));
  }
}
void fft(vector<int> &a, int pw[], int lg) {
  for (int i = 0; i < a.size(); ++i) {
    int r = (rev[i] >> (max_lev - lg)) & (a.size() - 1);
    if (i < r) {
      swap(a[i], a[r]);
    }
  }
  int pl = max_n / 2;
  for (int st = 1; st < a.size(); st *= 2) {
    for (int i = 0; i < a.size(); i += 2 * st) {
      int cur = 0;
      for (int j = i; j < i + st; ++j) {
        int y = mul(pw[cur], a[j + st]);
        a[j + st] = (a[j] + mod - y) % mod;
        a[j] = (a[j] + y) % mod;
        cur += pl;
      }
    }
    pl /= 2;
  }
}
const int max_triv = 20;
long long res[max_triv];
vector<int> triv_product(vector<int> &a, vector<int> &b) {
  for (int i = 0; i + 1 < a.size() + b.size(); ++i) {
    res[i] = 0;
  }
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < b.size(); ++j) {
      res[i + j] += 1LL * a[i] * b[j];
    }
  }
  vector<int> v(a.size() + b.size() - 1);
  for (int i = 0; i < v.size(); ++i) {
    v[i] = res[i] % mod;
  }
  return v;
}
vector<int> product(vector<int> &a, vector<int> &b) {
  if (a.size() + b.size() < max_triv) {
    return triv_product(a, b);
  }
  if (pw[0] == 0) {
    puts("NTT was not initialized");
    exit(228);
  }
  int sz = a.size() + b.size() - 1;
  int N = 1, lg = 0;
  while (N < sz) {
    N *= 2;
    ++lg;
  }
  while (a.size() < N) {
    a.push_back(0);
  }
  while (b.size() < N) {
    b.push_back(0);
  }
  fft(a, pw, lg);
  fft(b, pw, lg);
  for (int i = 0; i < a.size(); ++i) {
    a[i] = mul(a[i], b[i]);
  }
  fft(a, rpw, lg);
  int rn = power(N, mod - 2);
  vector<int> ans(sz);
  for (int i = 0; i < sz; ++i) {
    ans[i] = mul(a[i], rn);
  }
  return ans;
}
}  // namespace ntt
const int max_n = 20222, mod = 998244353;
int n, k, q, A[max_n], a[max_n];
int dp[max_n];
vector<int> solve(int l, int r) {
  if (l == r) {
    return {1, a[r]};
  }
  int mid = (l + r) / 2;
  vector<int> a = solve(l, mid);
  vector<int> b = solve(mid + 1, r);
  return ntt::product(a, b);
}
int main() {
  ntt::init();
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
  }
  scanf("%d", &q);
  while (q--) {
    copy(A, A + n, a);
    int tp, x;
    scanf("%d%d", &tp, &x);
    if (tp == 1) {
      int pos, d;
      scanf("%d%d", &pos, &d);
      --pos;
      a[pos] = d;
    } else {
      int l, r, d;
      scanf("%d%d%d", &l, &r, &d);
      --l;
      for (int i = l; i < r; ++i) {
        a[i] += d;
      }
    }
    for (int i = 0; i < n; ++i) {
      a[i] = ((x - a[i]) % mod + mod) % mod;
    }
    vector<int> ans = solve(0, n - 1);
    printf("%d\n", ans[k]);
  }
  return 0;
}