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 = (int)4e4 + 15;
const int inf = (long long)0x3f3f3f3f;
const int MOD = 998244353;
int a[N], b[N];
inline int addMod(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
inline int quickPow(int a, int b) {
  int ans = 1, base = a;
  while (b) {
    if (b & 1) {
      ans = (long long)ans * base % MOD;
    }
    base = (long long)base * base % MOD;
    b >>= 1;
  }
  return ans;
}
void change(vector<int>& y, int len) {
  for (int i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) {
      swap(y[i], y[j]);
    }
    int k = len / 2;
    while (j >= k) {
      j = j - k;
      k = k / 2;
    }
    if (j < k) {
      j += k;
    }
  }
}
void ntt(vector<int>& y, int len, int on) {
  change(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    int gn = quickPow(3, (MOD - 1) / h);
    for (int j = 0; j < len; j += h) {
      int g = 1;
      for (int k = j; k < j + h / 2; k++) {
        int u = y[k];
        int t = (long long)g * y[k + h / 2] % MOD;
        y[k] = addMod(u, t);
        y[k + h / 2] = addMod(u, MOD - t);
        g = (long long)g * gn % MOD;
      }
    }
  }
  if (on == -1) {
    reverse(y.begin() + 1, y.begin() + len);
    for (int i = 0, inv = quickPow(len, MOD - 2); i < len; i++) {
      y[i] = (long long)y[i] * inv % MOD;
    }
  }
}
inline void mul(vector<int> a, vector<int> b, vector<int>& res, int len,
                int k) {
  ntt(a, len, 1);
  ntt(b, len, 1);
  for (int i = 0; i < len; i++) {
    res[i] = (long long)a[i] * b[i] % MOD;
  }
  ntt(res, len, -1);
}
inline vector<int> mul(int l, int r, int k) {
  if (l == r) {
    vector<int> ret;
    ret.push_back(1);
    ret.push_back(b[l]);
    return ret;
  }
  int m = (l + r) >> 1;
  vector<int> t1 = mul(l, m, k);
  vector<int> t2 = mul(m + 1, r, k);
  vector<int> t3;
  int tl = t1.size() * 2;
  t1.resize(tl);
  t2.resize(tl);
  t3.resize(tl);
  mul(t1, t2, t3, tl, k);
  return t3;
}
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int q;
    scanf("%d", &q);
    while (q--) {
      int op, q;
      scanf("%d%d", &op, &q);
      for (int i = 1; i <= n; i++) {
        b[i] = addMod(q, MOD - a[i]);
      }
      if (op == 1) {
        int i, d;
        scanf("%d%d", &i, &d);
        b[i] = addMod(q, MOD - d);
      } else {
        int l, r, d;
        scanf("%d%d%d", &l, &r, &d);
        for (int i = l; i <= r; i++) {
          b[i] = addMod(b[i], MOD - d);
        }
      }
      vector<int> res = mul(1, n, k);
      printf("%d\n", res[k]);
    }
  }
}
