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;
struct nd {
  int pma[32];
  nd() {
    for (int i = 0; i < 32; i++) pma[i] = -(1 << 29);
  }
};
int n, k;
const int maxn = 200200;
nd t[maxn << 1];
nd merge(nd &a, nd &b) {
  nd ans;
  for (int i = 0; i < 1 << k; i++) ans.pma[i] = max(a.pma[i], b.pma[i]);
  return ans;
}
void modify(int p, int u[]) {
  nd x;
  for (int i = 0; i < 1 << k; i++) {
    int s = 0;
    for (int j = 0; j < k; j++) {
      if (i & (1 << j))
        s += u[j];
      else
        s -= u[j];
    }
    x.pma[i] = s;
  }
  for (t[p += n] = x; p > 1; p >>= 1) t[p >> 1] = merge(t[p], t[p ^ 1]);
}
nd query(int l, int r) {
  nd res;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = merge(res, t[l++]);
    if (r & 1) res = merge(res, t[--r]);
  }
  return res;
}
int v[10];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) scanf("%d", v + j);
    modify(i, v);
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 2) {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      nd p = query(l, r);
      int ans = -(1 << 30);
      int msk = (1 << k) - 1;
      for (int j = 0; j < 1 << k; j++)
        ans = max(ans, p.pma[j] + p.pma[j ^ msk]);
      printf("%d\n", ans);
    } else {
      int p;
      scanf("%d", &p);
      p--;
      for (int j = 0; j < k; j++) scanf("%d", v + j);
      modify(p, v);
    }
  }
  return 0;
}
