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 + 7;
int n, k;
class Aint {
  pair<int, int> t[4 * N];
  pair<int, int> ans;
  int l, r, val;
  void dfs(int nod = 1, int st = 1, int dr = n) {
    if (l <= st && dr <= r) {
      t[nod] = {val, val};
      return;
    }
    int mij((st + dr) >> 1);
    nod <<= 1;
    if (l <= mij) dfs(nod, st, mij);
    if (r > mij) dfs(nod + 1, mij + 1, dr);
    t[nod >> 1] = {min(t[nod].first, t[nod + 1].first),
                   max(t[nod].second, t[nod + 1].second)};
  }
  void run(int nod = 1, int st = 1, int dr = n) {
    if (l <= st && dr <= r) {
      ans = {min(ans.first, t[nod].first), max(ans.second, t[nod].second)};
      return;
    }
    int mij((st + dr) >> 1);
    nod <<= 1;
    if (l <= mij) run(nod, st, mij);
    if (r > mij) run(nod + 1, mij + 1, dr);
  }

 public:
  Aint() {
    for (int i = 0; i < 4 * N; ++i) t[i] = {1e7, -1e7};
  }
  void update(int aval, int poz) {
    val = aval;
    l = r = poz;
    dfs();
  }
  int query(int ast, int adr) {
    l = ast;
    r = adr;
    ans = {1e7, -1e7};
    run();
    return ans.second - ans.first;
  }
} aint[35];
int pzu;
int v[6];
void bkt(int p = 0, int x = 0, int val = 0) {
  if (p == k) {
    aint[x].update(val, pzu);
    return;
  }
  bkt(p + 1, x, val - v[p]);
  bkt(p + 1, (x ^ (1 << p)), val + v[p]);
}
int main() {
  int q, t, l, r;
  cin >> n >> k;
  for (pzu = 1; pzu <= n; ++pzu) {
    for (int j = 0; j < k; ++j) cin >> v[j];
    bkt();
  }
  cin >> q;
  while (q--) {
    cin >> t;
    if (t - 1) {
      cin >> l >> r;
      int ans(0);
      for (int i = 0; i < (1 << k); ++i) ans = max(ans, aint[i].query(l, r));
      cout << ans << '\n';
    } else {
      cin >> pzu;
      for (int i = 0; i < k; ++i) cin >> v[i];
      bkt();
    }
  }
  return 0;
}
