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;
#pragma optimize("-O3")
int ti[800005][32];
int ta[800005][32];
int maxi(int v, int tl, int tr, int l, int r, int i) {
  if (l > r) return -1000000000;
  if (l == tl && r == tr) {
    return ta[v][i];
  }
  int tm = (tl + tr) / 2;
  return max(maxi(v * 2, tl, tm, l, min(r, tm), i),
             maxi(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, i));
}
void updatea(int v, int tl, int tr, int pos, int new_val, int i) {
  if (tl == tr) {
    ta[v][i] = new_val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      updatea(v * 2, tl, tm, pos, new_val, i);
    else
      updatea(v * 2 + 1, tm + 1, tr, pos, new_val, i);
    ta[v][i] = max(ta[v * 2][i], ta[v * 2 + 1][i]);
  }
}
int mini(int v, int tl, int tr, int l, int r, int i) {
  if (l > r) return 1000000000;
  if (l == tl && r == tr) {
    return ti[v][i];
  }
  int tm = (tl + tr) / 2;
  return min(mini(v * 2, tl, tm, l, min(r, tm), i),
             mini(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, i));
}
void updatei(int v, int tl, int tr, int pos, int new_val, int i) {
  if (tl == tr) {
    ti[v][i] = new_val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      updatei(v * 2, tl, tm, pos, new_val, i);
    else
      updatei(v * 2 + 1, tm + 1, tr, pos, new_val, i);
    ti[v][i] = min(ti[v * 2][i], ti[v * 2 + 1][i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int arr[n][k];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      cin >> arr[i][j];
    }
    for (int j = 0; j < (1 << k); j++) {
      int val = 0;
      for (int ind = 0; ind < k; ind++) {
        if (((1 << ind) & j) != 0) {
          val += arr[i][ind];
        } else {
          val -= arr[i][ind];
        }
      }
      updatea(1, 0, n - 1, i, val, j);
      updatei(1, 0, n - 1, i, val, j);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      x--;
      int vals[k];
      for (int i = 0; i < k; i++) {
        cin >> vals[i];
      }
      for (int j = 0; j < (1 << k); j++) {
        int sum = 0;
        for (int ind = 0; ind < k; ind++) {
          if (((1 << ind) & j) != 0) {
            sum += vals[ind];
          } else {
            sum -= vals[ind];
          }
        }
        updatea(1, 0, n - 1, x, sum, j);
        updatei(1, 0, n - 1, x, sum, j);
      }
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      int maxim = 0;
      for (int i = 0; i < (1 << k); i++) {
        maxim =
            max(maxim, maxi(1, 0, n - 1, l, r, i) - mini(1, 0, n - 1, l, r, i));
      }
      cout << maxim << endl;
    }
  }
  return 0;
}
