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 n;
struct seg {
  int t[N << 1][2];
  void modify(int pos, int val) {
    pos += n;
    for (t[pos][0] = t[pos][1] = val; pos != 1; pos >>= 1) {
      t[pos >> 1][0] = min(t[pos][0], t[pos ^ 1][0]);
      t[pos >> 1][1] = max(t[pos][1], t[pos ^ 1][1]);
    }
  }
  int query(int l, int r, bool ismx) {
    int resp = 1000000007;
    if (ismx) resp = -resp;
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) {
        if (ismx)
          resp = max(resp, t[l++][1]);
        else
          resp = min(resp, t[l++][0]);
      }
      if (r & 1) {
        if (ismx)
          resp = max(resp, t[--r][1]);
        else
          resp = min(resp, t[--r][0]);
      }
    }
    return resp;
  }
} s[16];
int p[10];
int main() {
  int k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &p[j]);
    }
    for (int j = 0; j < 16; j++) {
      int val = 0;
      for (int g = 0; g < 5; g++) {
        if (j & (1 << g))
          val -= p[5 - g - 1];
        else
          val += p[5 - g - 1];
      }
      s[j].modify(i, val);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int i;
      scanf("%d", &i);
      i--;
      for (int j = 0; j < k; j++) {
        scanf("%d", &p[j]);
      }
      for (int j = 0; j < 16; j++) {
        int val = 0;
        for (int g = 0; g < 5; g++) {
          if (j & (1 << g))
            val -= p[5 - g - 1];
          else
            val += p[5 - g - 1];
        }
        s[j].modify(i, val);
      }
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      int best = 0;
      for (int j = 0; j < 16; j++) {
        int mx = s[j].query(l, r, true), mn = s[j].query(l, r, false);
        best = max(best, mx - mn);
      }
      printf("%d\n", best);
    }
  }
  return 0;
}
