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;
struct Tree {
  static const int LOW = INT_MIN;
  int f(int a, int b) { return max(a, b); }
  vector<int> s;
  int n;
  Tree(int n = 0, int def = 0) : s(2 * n, def), n(n) {}
  void update(int pos, int val) {
    for (s[pos += n] = val; pos > 1; pos /= 2)
      s[pos / 2] = f(s[pos & ~1], s[pos | 1]);
  }
  int query(int b, int e) {
    int ra = LOW, rb = LOW;
    for (b += n, e += n; b < e; b /= 2, e /= 2) {
      if (b % 2) ra = f(ra, s[b++]);
      if (e % 2) rb = f(s[--e], rb);
    }
    return f(ra, rb);
  }
};
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int n, k;
  cin >> n >> k;
  int a[n][k];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      cin >> a[i][j];
    }
  }
  vector<Tree> st((1 << k), Tree(n));
  for (int i = 0; i < (1 << k); i++) {
    st[i] = Tree(n);
    for (int j = 0; j < n; j++) {
      int ans = 0;
      for (int ii = 0; ii < k; ii++) {
        if (i & (1 << ii))
          ans += a[j][ii];
        else
          ans -= a[j][ii];
      }
      st[i].update(j, ans);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int idx;
      cin >> idx;
      idx--;
      int b[k];
      for (int i = 0; i < k; i++) cin >> b[i];
      int ans = 0;
      for (int i = 0; i < (1 << k); i++) {
        ans = 0;
        for (int ii = 0; ii < k; ii++) {
          if (i & (1 << ii))
            ans += b[ii];
          else
            ans -= b[ii];
        }
        st[i].update(idx, ans);
      }
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      int ans = 0;
      for (int i = 0; i < (1 << k); i++) {
        int a = st[i].query(l, r);
        int b = st[(1 << k) - 1 - i].query(l, r);
        ans = max(ans, a + b);
      }
      cout << ans << endl;
    }
  }
  return 0;
}
