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;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  x = (x > y ? y : x);
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  x = (x < y ? y : x);
}
using ll = long long;
const string FILENAME = "input";
const int MAXN = 20228;
const ll MOD = 998244353;
int rev[300000 * 4];
const int G = 3;
int mul(int a, int b) { return (1LL * a * b) % MOD; }
int sum(int a, int b) { return (1LL * a + b + MOD) % MOD; }
int powm(int a, int b) {
  int cur = a;
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, cur);
    }
    cur = mul(cur, cur);
    b >>= 1;
  }
  return res;
}
void FFT(vector<ll> &a, int n, ll op) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1; i < n; i <<= 1) {
    ll wn = powm(G, (MOD - 1) / i >> 1);
    if (op) wn = powm(wn, MOD - 2);
    for (int j = 0; j < n; j += i << 1) {
      ll w = 1;
      for (int k = j; k < j + i; k++, w = 1ll * w * wn % MOD) {
        ll x = a[k], y = 1ll * w * a[k + i] % MOD;
        a[k] = (x + y) % MOD;
        a[k + i] = (x - y + MOD) % MOD;
      }
    }
  }
  if (op) {
    ll invn = powm(n, MOD - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * invn % MOD;
  }
}
vector<ll> multiply(vector<ll> a, vector<ll> b) {
  int n = max((int)(a).size(), (int)(b).size());
  int ns = 1, len = 0, m = n * 2;
  for (; ns <= m; len++) ns <<= 1;
  for (int i = 1; i < ns; i++) {
    rev[i] = rev[i >> 1] >> 1 ^ ((i & 1) << (len - 1));
  }
  a.resize(ns);
  b.resize(ns);
  FFT(a, ns, 0);
  FFT(b, ns, 0);
  for (int i = 0; i < ns; i++) a[i] = mul(a[i], b[i]);
  FFT(a, ns, 1);
  return a;
}
int n, k;
ll a[MAXN];
ll b[MAXN];
vector<ll> get(int l, int r) {
  if (l == r) {
    vector<ll> res;
    res.resize(2);
    res[0] = 1;
    res[1] = b[l];
    return res;
  }
  vector<ll> f = get(l, (l + r) / 2);
  vector<ll> f1 = get((l + r) / 2 + 1, r);
  vector<ll> g = multiply(f, f1);
  while (!g.empty() && g.back() == 0) {
    g.pop_back();
  }
  return g;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int q;
  cin >> q;
  for (int it = 0; it < q; it++) {
    int t;
    cin >> t;
    if (t == 1) {
      int q, i, d;
      cin >> q >> i >> d;
      int c = a[i];
      a[i] = d;
      for (int i = 1; i <= n; i++) {
        b[i] = (q - a[i]);
        b[i] %= MOD;
        if (b[i] < 0) {
          b[i] += MOD;
        }
      }
      vector<ll> kek = get(1, n);
      a[i] = c;
      cout << kek[k] << '\n';
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      for (int i = l; i <= r; i++) {
        a[i] += d;
      }
      for (int i = 1; i <= n; i++) {
        b[i] = (q - a[i]);
        b[i] %= MOD;
        if (b[i] < 0) {
          b[i] += MOD;
        }
      }
      vector<ll> kek = get(1, n);
      for (int i = l; i <= r; i++) {
        a[i] -= d;
      }
      cout << kek[k] << '\n';
    }
  }
  return 0;
}