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 n, k, IT[1000000][32];
void update(int x, vector<int> p, int i, int l, int r) {
  if (x < l || r < x) return;
  if (l == r) {
    for (int m = 0; m < (1 << k); ++m) {
      int calc = 0;
      for (int i = 0; i < k; ++i) (m & (1 << i)) ? calc += p[i] : calc -= p[i];
      IT[i][m] = calc;
    }
    return;
  }
  update(x, p, 2 * i, l, (l + r) / 2);
  update(x, p, 2 * i + 1, (l + r) / 2 + 1, r);
  for (int m = 0; m < (1 << k); ++m)
    IT[i][m] = max(IT[2 * i][m], IT[2 * i + 1][m]);
}
int get(int x, int y, int m, int i, int l, int r) {
  if (y < l || r < x) return -1e9;
  if (x <= l && r <= y) return IT[i][m];
  int x1 = get(x, y, m, 2 * i, l, (l + r) / 2);
  int x2 = get(x, y, m, 2 * i + 1, (l + r) / 2 + 1, r);
  return max(x1, x2);
}
int ans(int l, int r) {
  int res = 0;
  for (int m = 0, x1, x2; m < (1 << k); ++m)
    x1 = get(l, r, m, 1, 1, n), x2 = get(l, r, ((1 << k) - 1) ^ m, 1, 1, n),
    res = max(res, x1 + x2);
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    vector<int> p;
    for (int i = 0, x; i < k; ++i) cin >> x, p.push_back(x);
    update(i, p, 1, 1, n);
    p.clear();
  }
  int q;
  cin >> q;
  for (int t, i, l, r; q--;) {
    cin >> t;
    if (t == 1) {
      cin >> i;
      vector<int> p;
      for (int i = 0, x; i < k; ++i) cin >> x, p.push_back(x);
      update(i, p, 1, 1, n);
      p.clear();
    } else
      cin >> l >> r, cout << ans(l, r) << '\n';
  }
}
