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;
const int N = 2e5 + 5;
int MOD = 1e9 + 7;
int n, q, k;
int a[N][10], t[4 * N][32];
void update(int id, int x, int y, int l) {
  if (x > l or y < l) return;
  if (x == y) {
    for (int i = 0; i < (1 << (k - 1)); i++) {
      int v = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1) {
          v -= a[l][j + 1];
        } else {
          v += a[l][j + 1];
        }
      }
      t[id][i] = v;
      t[id][(i ^ ((1 << k) - 1))] = -v;
    }
    return;
  }
  int mid = (x + y) >> 1;
  update(2 * id, x, mid, l);
  update(2 * id + 1, mid + 1, y, l);
  for (int i = 0; i < (1 << k); i++) {
    t[id][i] = max(t[2 * id][i], t[2 * id + 1][i]);
  }
}
int query(int id, int bit, int x, int y, int l, int r) {
  if (l > y or r < x) return -MOD;
  if (l <= x and r >= y) return t[id][bit];
  int mid = (x + y) >> 1;
  int ans = max(query(2 * id, bit, x, mid, l, r),
                query(2 * id + 1, bit, mid + 1, y, l, r));
  return ans;
}
signed main() {
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 1; i <= 4 * n; i++) {
    for (int j = 0; j < (1 << k); j++) t[i][j] = -MOD;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      scanf("%d", &a[i][j]);
    }
    update(1, 1, n, i);
  }
  scanf("%d", &q);
  int ty, ind, l, r;
  while (q--) {
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d", &ind);
      for (int j = 1; j <= k; j++) {
        scanf("%d", &a[ind][j]);
      }
      update(1, 1, n, ind);
    } else {
      scanf("%d", &l);
      scanf("%d", &r);
      int ans = 0;
      for (int i = 0; i < (1 << (k - 1)); i++) {
        ans = max(ans, query(1, i, 1, n, l, r) +
                           query(1, (i ^ ((1 << k) - 1)), 1, n, l, r));
      }
      printf("%d\n", ans);
    }
  }
}
