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;
int n, k;
int a[200005][5];
int sgn[32][5];
int Max[32][800005], Min[32][800005];
void update(int p, int x, int v, int st = 1, int dr = n, int nod = 1) {
  if (st == dr) {
    Max[p][nod] = Min[p][nod] = v;
    return;
  }
  int mij = (st + dr) / 2;
  if (x <= mij)
    update(p, x, v, st, mij, nod * 2);
  else
    update(p, x, v, mij + 1, dr, nod * 2 + 1);
  Max[p][nod] = max(Max[p][nod * 2], Max[p][nod * 2 + 1]);
  Min[p][nod] = min(Min[p][nod * 2], Min[p][nod * 2 + 1]);
}
pair<int, int> query(int p, int x, int y, int st = 1, int dr = n, int nod = 1) {
  if (x <= st && dr <= y) return {Min[p][nod], Max[p][nod]};
  int mij = (st + dr) / 2;
  pair<int, int> p1, p2;
  p1 = p2 = {2000000000, -2000000000};
  if (x <= mij) p1 = query(p, x, y, st, mij, nod * 2);
  if (mij + 1 <= y) p2 = query(p, x, y, mij + 1, dr, nod * 2 + 1);
  return {min(p1.first, p2.first), max(p1.second, p2.second)};
}
int main() {
  scanf("%d%d", &n, &k);
  int K = (1 << k);
  for (int p = 0; p < K; ++p) {
    for (int i = 0; i < k; ++i) {
      if ((1 << i) & p)
        sgn[p][i] = 1;
      else
        sgn[p][i] = -1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) scanf("%d", &a[i][j]);
    for (int p = 0; p < K; ++p) {
      int Sum = 0;
      for (int j = 0; j < k; ++j) Sum = Sum + a[i][j] * sgn[p][j];
      update(p, i, Sum);
    }
  }
  int q, tip, x, y, b[5];
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &tip);
    if (tip == 1) {
      scanf("%d", &x);
      for (int i = 0; i < k; ++i) scanf("%d", &b[i]);
      int Sum = 0;
      for (int p = 0; p < K; ++p) {
        int Sum = 0;
        for (int j = 0; j < k; ++j) Sum = Sum + b[j] * sgn[p][j];
        update(p, x, Sum);
      }
    } else {
      scanf("%d%d", &x, &y);
      int Sol = 0;
      for (int p = 0; p < K; ++p) {
        pair<int, int> ans = query(p, x, y);
        Sol = max(Sol, ans.second - ans.first);
      }
      printf("%d\n", Sol);
    }
  }
  return 0;
}
