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 sege {
 private:
  sege *son[2];
  int v;
  sege *s(int i) { return son[i] ? son[i] : son[i] = new sege; }

 public:
  sege() { son[0] = son[1] = NULL, v = -1e9; }
  void edi(int l, int r, int p, int v) {
    if (l == r)
      this->v = v;
    else {
      if (p <= (l + r) / 2)
        s(0)->edi(l, (l + r) / 2, p, v);
      else
        s(1)->edi((l + r) / 2 + 1, r, p, v);
      this->v = max(s(0)->v, s(1)->v);
    }
  }
  int getv(int l, int r, int ql, int qr) {
    if (l == ql && r == qr) return v;
    if (qr <= (l + r) / 2) return s(0)->getv(l, (l + r) / 2, ql, qr);
    if (ql > (l + r) / 2) return s(1)->getv((l + r) / 2 + 1, r, ql, qr);
    return max(s(0)->getv(l, (l + r) / 2, ql, (l + r) / 2),
               s(1)->getv((l + r) / 2 + 1, r, (l + r) / 2 + 1, qr));
  }
} rt[37];
int n, k;
inline void refre(int i) {
  int b[7];
  for (int i = 0; i < k; ++i) scanf("%d", b + i);
  for (int j = 0; j < (1 << k); ++j) {
    int res = 0;
    for (int i = 0; i < k; ++i) res += j & (1 << i) ? b[i] : -b[i];
    rt[j].edi(1, n, i, res);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) refre(i);
  int t, a, b;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &a, &b);
    if (a == 1)
      refre(b);
    else {
      int ans = 0;
      scanf("%d", &a);
      for (int j = 0; j < (1 << (k - 1)); ++j) {
        ans = max(ans, rt[j].getv(1, n, b, a) +
                           rt[j ^ ((1 << k) - 1)].getv(1, n, b, a));
      }
      cout << ans << '\n';
    }
  }
}
