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 = 200005;
array<long long, 32> st[2 * N];
int n, k;
array<long long, 32> merge(array<long long, 32> a, array<long long, 32> b) {
  array<long long, 32> ans;
  for (int i = 0; i < (1 << k); i++) ans[i] = max(a[i], b[i]);
  return ans;
}
long long query(int l, int r) {
  array<long long, 32> ans;
  fill(ans.begin(), ans.end(), -(long long)1e18);
  for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans = merge(ans, st[l++]);
    if (r & 1) ans = merge(st[--r], ans);
  }
  long long ret = 0;
  int inv = (1 << k) - 1;
  for (int i = 0; i < (1 << k); i++) ret = max(ret, ans[i] + ans[inv ^ i]);
  return ret;
}
void build() {
  for (int p = n - 1; p > 0; p--) st[p] = merge(st[2 * p], st[2 * p + 1]);
}
void update(int p, array<long long, 32> val) {
  st[p += n] = val;
  while (p >>= 1) st[p] = merge(st[2 * p], st[2 * p + 1]);
}
array<long long, 32> read() {
  static long long vet[6];
  array<long long, 32> ret;
  fill(ret.begin(), ret.end(), 0);
  for (int j = 0; j < k; j++) scanf("%lld", vet + j);
  for (int j = 0; j < (1 << k); j++) {
    for (int q = 0; q < k; q++) {
      if (j & (1 << q))
        ret[j] += vet[q];
      else
        ret[j] -= vet[q];
    }
  }
  return ret;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) st[n + i] = read();
  build();
  int q;
  scanf("%d", &q);
  int t, l, r;
  while (q--) {
    scanf("%d %d", &t, &l);
    if (t == 1) {
      update(l - 1, read());
    } else {
      scanf("%d", &r);
      printf("%lld\n", query(l - 1, r - 1));
    }
  }
}
