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 T {
  int t[524300], L, R;
  void upd(int x, int k) {
    k += 262143;
    t[k] = x;
    k >>= 1;
    while (k) t[k] = max(t[k << 1], t[(k << 1) ^ 1]), k >>= 1;
  }
  int get(int k, int l, int r) {
    if (l >= L && r <= R) return t[k];
    int m = (l + r) >> 1, lc = k << 1, rc = lc ^ 1;
    if (L > m) return get(rc, m + 1, r);
    if (R <= m) return get(lc, l, m);
    return max(get(lc, l, m), get(rc, m + 1, r));
  }
  int qry(int l, int r) {
    L = l, R = r;
    return get(1, 1, 262144);
  }
  void bld() {
    for (int i = 262143; i; --i) t[i] = max(t[i << 1], t[(i << 1) ^ 1]);
  }
} t[32];
int n, k, q, a[200005][5], res[32];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) scanf("%d", a[i] + j);
  }
  for (int _ = 0; _ < (1 << k); ++_) {
    for (int i = 1; i <= n; ++i) {
      int o = 0;
      for (int j = 0; j < k; ++j) {
        if (_ & (1 << j))
          o += a[i][j];
        else
          o -= a[i][j];
      }
      t[_].t[i + 262143] = o;
    }
    for (int i = n + 262144; i <= 524287; ++i) t[_].t[i] = -1000000007;
    t[_].bld();
  }
  scanf("%d", &q);
  while (q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int x;
      scanf("%d", &x);
      for (int i = 0; i < k; ++i) scanf("%d", a[x] + i);
      for (int _ = 0; _ < (1 << k); ++_) {
        int o = 0;
        for (int i = 0; i < k; ++i) {
          if (_ & (1 << i))
            o += a[x][i];
          else
            o -= a[x][i];
        }
        t[_].upd(o, x);
      }
    } else {
      int l, r, _ = (1 << k) - 1, ans = 0;
      scanf("%d%d", &l, &r);
      for (int i = 0; i <= _; ++i) res[i] = t[i].qry(l, r);
      for (int i = 0; i <= _; ++i) ans = max(ans, res[i] + res[_ ^ i]);
      printf("%d\n", ans);
    }
  }
  return 0;
}
