You are given an array a of n points in k-dimensional space. Let the distance between two points a_x and a_y be ∑ _{i = 1}^{k} |a_{x, i} - a_{y, i}| (it is also known as Manhattan distance).

You have to process q queries of the following two types:

  * 1 i b_1 b_2 ... b_k — set i-th element of a to the point (b_1, b_2, ..., b_k);
  * 2 l r — find the maximum distance between two points a_i and a_j, where l ≤ i, j ≤ r.

Input

The first line contains two numbers n and k (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 5) — the number of elements in a and the number of dimensions of the space, respectively.

Then n lines follow, each containing k integers a_{i, 1}, a_{i, 2}, ..., a_{i, k} (-10^6 ≤ a_{i, j} ≤ 10^6) — the coordinates of i-th point.

The next line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, each denoting a query. There are two types of queries:

  * 1 i b_1 b_2 ... b_k (1 ≤ i ≤ n, -10^6 ≤ b_j ≤ 10^6) — set i-th element of a to the point (b_1, b_2, ..., b_k);
  * 2 l r (1 ≤ l ≤ r ≤ n) — find the maximum distance between two points a_i and a_j, where l ≤ i, j ≤ r.



There is at least one query of the second type.

Output

Print the answer for each query of the second type.

Example

Input


5 2
1 2
2 3
3 4
4 5
5 6
7
2 1 5
2 1 3
2 3 5
1 5 -1 -2
2 1 5
1 4 -1 -2
2 1 5


Output


8
4
4
12
10
#include <bits/stdc++.h>
using namespace std;
char cb[1 << 15], *cs = cb, *ct = cb;
inline void read(int &res) {
  char ch;
  bool flag = 0;
  for (; !isdigit(
           ch = (cs == ct && (ct = (cs = cb) + fread(cb, 1, 1 << 15, stdin),
                             cs == ct)
                     ? 0
                     : *cs++));)
    if (ch == '-') flag = 1;
  for (res = ch - '0';
       isdigit(ch = (cs == ct && (ct = (cs = cb) + fread(cb, 1, 1 << 15, stdin),
                                 cs == ct)
                         ? 0
                         : *cs++));
       res = res * 10 + ch - '0')
    ;
  (flag) && (res = -res);
}
int n, x[800005][5], k;
int Max[800005][32], Min[800005][32];
void Insert(int now, int l, int r, int pos, int *a) {
  if (l > pos || r < pos) return;
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) Min[now][i] = Max[now][i] = a[i];
    return;
  }
  int mid = (l + r) >> 1;
  Insert(now << 1, l, mid, pos, a), Insert(now << 1 | 1, mid + 1, r, pos, a);
  for (int i = 0; i < (1 << k); i++)
    Min[now][i] = min(Min[now << 1][i], Min[now << 1 | 1][i]),
    Max[now][i] = max(Max[now << 1][i], Max[now << 1 | 1][i]);
}
void Qmin(int now, int l, int r, int ql, int qr, int *ret) {
  if (l > qr || r < ql) return;
  if (ql <= l && r <= qr) {
    for (int i = 0; i < (1 << k); i++) ret[i] = min(ret[i], Min[now][i]);
    return;
  }
  int mid = (l + r) >> 1;
  Qmin(now << 1, l, mid, ql, qr, ret),
      Qmin(now << 1 | 1, mid + 1, r, ql, qr, ret);
}
void Qmax(int now, int l, int r, int ql, int qr, int *ret) {
  if (l > qr || r < ql) return;
  if (ql <= l && r <= qr) {
    for (int i = 0; i < (1 << k); i++) ret[i] = max(ret[i], Max[now][i]);
    return;
  }
  int mid = (l + r) >> 1;
  Qmax(now << 1, l, mid, ql, qr, ret),
      Qmax(now << 1 | 1, mid + 1, r, ql, qr, ret);
}
int calc(int sta, int loc) {
  int ret = 0;
  for (int i = 0; i < k; i++)
    if ((sta >> i) & 1)
      ret += x[loc][i];
    else
      ret -= x[loc][i];
  return ret;
}
int sta[2][32];
int main() {
  read(n), read(k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) read(x[i][j]);
    for (int p = 0; p < (1 << k); p++) sta[0][p] = calc(p, i);
    Insert(1, 1, n, i, sta[0]);
  }
  int q;
  for (read(q); q--;) {
    int op;
    read(op);
    if (op == 1) {
      int i;
      read(i);
      for (int j = 0; j < k; j++) read(x[i][j]);
      for (int p = 0; p < (1 << k); p++) sta[0][p] = calc(p, i);
      Insert(1, 1, n, i, sta[0]);
    } else {
      int l, r;
      read(l), read(r);
      int ans = 0;
      memset(sta[0], -0x3f, sizeof sta[0]);
      memset(sta[1], 0x3f, sizeof sta[1]);
      Qmax(1, 1, n, l, r, sta[0]), Qmin(1, 1, n, l, r, sta[1]);
      for (int i = 0; i < (1 << k); i++) ans = max(sta[0][i] - sta[1][i], ans);
      printf("%d\n", ans);
    }
  }
}
