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 = 2e5 + 5;
int ma[N << 2][1 << 4], mi[N << 2][1 << 4], k;
int p[N][5];
void get(int T, int ind) {
  for (int i = 0; i < 1 << k; i++) {
    int sum = p[ind][0];
    for (int j = 0; j < k; j++) {
      if (i & (1 << j))
        sum += p[ind][j + 1];
      else
        sum -= p[ind][j + 1];
    }
    ma[T][i] = mi[T][i] = sum;
  }
}
void build(int i, int l, int r) {
  if (l == r) return get(i, l);
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  for (int j = 0; j < 1 << k; j++) {
    ma[i][j] = max(ma[i << 1][j], ma[i << 1 | 1][j]);
    mi[i][j] = min(mi[i << 1][j], mi[i << 1 | 1][j]);
  }
}
void update(int i, int l, int r, int a) {
  if (l > a || r < a) return;
  if (l == r) return get(i, a);
  int mid = (l + r) >> 1;
  update(i << 1, l, mid, a);
  update(i << 1 | 1, mid + 1, r, a);
  for (int j = 0; j < 1 << k; j++) {
    ma[i][j] = max(ma[i << 1][j], ma[i << 1 | 1][j]);
    mi[i][j] = min(mi[i << 1][j], mi[i << 1 | 1][j]);
  }
}
vector<int> ans, ans2;
void solve(int i, int l, int r, int a, int b) {
  if (l > b || r < a) return;
  if (l >= a && r <= b) {
    for (int j = 0; j < 1 << k; j++) {
      ans[j] = max(ans[j], ma[i][j]);
      ans2[j] = min(ans2[j], mi[i][j]);
    }
    return;
  }
  int mid = (l + r) >> 1;
  solve(i << 1, l, mid, a, b);
  solve(i << 1 | 1, mid + 1, r, a, b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < k; j++) cin >> p[i][j];
  k--;
  build(1, 0, n - 1);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int ind;
      cin >> ind;
      ind--;
      for (int i = 0; i < k + 1; i++) cin >> p[ind][i];
      update(1, 0, n - 1, ind);
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      ans.clear(), ans2.clear();
      ans.resize(1 << k, -1e9), ans2.resize(1 << k, 1e9);
      solve(1, 0, n - 1, l, r);
      int res = 0;
      for (int i = 0; i < 1 << k; i++) res = max(res, ans[i] - ans2[i]);
      cout << res << '\n';
    }
  }
  return 0;
}
