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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool smin(T &a, const T &b) {
  return a > b ? a = b : a;
}
template <typename T>
inline bool smax(T &a, const T &b) {
  return a < b ? a = b : a;
}
const int N = (int)1 << 15, mod = (int)998244353;
int a[N], b[N], dp[N];
int pw(int a, int b) {
  return b != 0 ? (long long)pw((long long)a * a % mod, b >> 1) *
                      (b & 1 ? a : 1) % mod
                : 1;
}
void NTT(int a[], int lg, bool inverse) {
  int N = 1 << lg;
  for (int i = 0; i < N; ++i) {
    int rev = 0;
    for (int j = 0; j < lg; ++j)
      if (i >> j & 1) rev |= 1 << (lg - j - 1);
    if (rev > i) swap(a[rev], a[i]);
  }
  for (int len = 2; len <= N; len <<= 1) {
    int wn = pw(3, (mod - 1) / len);
    if (inverse) wn = pw(wn, mod - 2);
    for (int j = 0; j < N; j += len) {
      int w = 1;
      for (int first = j, second = j + len / 2; second < j + len;
           ++first, ++second) {
        int u = a[first], v = (long long)a[second] * w % mod;
        a[first] = u + v;
        a[second] = u - v + mod;
        while (a[first] >= mod) a[first] -= mod;
        while (a[second] >= mod) a[second] -= mod;
        w = (long long)w * wn % mod;
      }
    }
  }
  if (inverse) {
    int div = pw(N, mod - 2);
    for (int i = 0; i < N; ++i) a[i] = (long long)a[i] * div % mod;
  }
}
int res[17][N], tmpa[17][N], tmpb[17][N];
void solve(int b, int e, int a[N], int d = 0) {
  if (b + 1 == e) {
    res[d][0] = 1;
    res[d][1] = a[b];
    return;
  }
  int m = b + e >> 1;
  solve(b, m, a, d + 1);
  for (int j = 0; j <= (m - b); ++j)
    tmpa[d][j] = res[d + 1][j], res[d + 1][j] = 0;
  solve(m, e, a, d + 1);
  for (int j = 0; j <= (e - m); ++j)
    tmpb[d][j] = res[d + 1][j], res[d + 1][j] = 0;
  int lg = 0;
  while ((1 << lg) <= e - b) ++lg;
  NTT(tmpa[d], lg, false);
  NTT(tmpb[d], lg, false);
  for (int j = 0; j < (1 << lg); ++j)
    res[d][j] = (long long)tmpa[d][j] * tmpb[d][j] % mod,
    tmpa[d][j] = tmpb[d][j] = 0;
  NTT(res[d], lg, true);
}
int32_t main() {
  int n, k;
  cin >> n >> k;
  for (int j = 0; j < n; ++j) cin >> a[j];
  int qm;
  cin >> qm;
  while (qm--) {
    int c;
    cin >> c;
    if (c == 1) {
      int q, p, d;
      cin >> q >> p >> d;
      --p;
      for (int j = 0; j < n; ++j) b[j] = q - a[j];
      b[p] = q - d;
    } else {
      int q, l, r, d;
      cin >> q >> l >> r >> d;
      --l;
      for (int j = 0; j < n; ++j) b[j] = q - a[j];
      for (int j = l; j < r; ++j) b[j] -= d;
    }
    for (int j = 0; j < n; ++j) b[j] = (b[j] % mod + mod) % mod;
    memset(res, 0, sizeof res);
    solve(0, n, b);
    cout << res[0][k] << '\n';
  }
}
