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;
int seg[800001][33][2];
int A[200001][5];
int k;
void change(int c, int l, int r, int pos) {
  if (l == r) {
    for (int mask = 0; mask < (1 << k); mask++) {
      int sum = 0;
      for (int j = 0; j < k; j++) sum += (mask & (1 << j) ? 1 : -1) * A[r][j];
      seg[c][mask][0] = sum;
      seg[c][mask][1] = sum;
    }
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    change(2 * c, l, m, pos);
  else
    change(2 * c + 1, m + 1, r, pos);
  for (int mask = 0; mask < (1 << k); mask++) {
    seg[c][mask][0] = min(seg[2 * c][mask][0], seg[2 * c + 1][mask][0]);
    seg[c][mask][1] = max(seg[2 * c][mask][1], seg[2 * c + 1][mask][1]);
  }
}
int Query(int c, int l, int r, int L, int R, int tp, int mask) {
  if (l > r || l > R || r < L) return (tp == 0 ? 666666666 : -666666666);
  if (l >= L && r <= R) return seg[c][mask][tp];
  int m = (l + r) / 2;
  if (tp == 0)
    return min(Query(2 * c, l, m, L, R, tp, mask),
               Query(2 * c + 1, m + 1, r, L, R, tp, mask));
  return max(Query(2 * c, l, m, L, R, tp, mask),
             Query(2 * c + 1, m + 1, r, L, R, tp, mask));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> k;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) cin >> A[i][j];
  for (int i = 1; i <= n; i++) change(1, 1, n, i);
  cin >> q;
  while (q--) {
    int tp, l, r, idx;
    cin >> tp;
    if (tp == 1) {
      cin >> idx;
      for (int j = 0; j < k; j++) cin >> A[idx][j];
      change(1, 1, n, idx);
    } else {
      cin >> l >> r;
      int ats = 0;
      for (int mask = 0; mask < (1 << k); mask++)
        ats = max(
            ats, Query(1, 1, n, l, r, 1, mask) - Query(1, 1, n, l, r, 0, mask));
      cout << ats << "\n";
    }
  }
}
