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;
void build(int node, int l, int r, int sign, vector<int>& st,
           vector<vector<int>>& a) {
  if (l == r) {
    st[node] = 0;
    for (int i = 0; i < k; ++i)
      if (sign & (1 << i))
        st[node] += a[l][i];
      else
        st[node] -= a[l][i];
  } else {
    int mid = (l + r) >> 1;
    build(node << 1, l, mid, sign, st, a);
    build(node << 1 | 1, mid + 1, r, sign, st, a);
    st[node] = max(st[node << 1], st[node << 1 | 1]);
  }
}
void update(int node, int l, int r, int pos, int sign, vector<int>& st,
            vector<int>& a) {
  if (l == r) {
    st[node] = 0;
    for (int i = 0; i < k; ++i)
      if (sign & (1 << i))
        st[node] += a[i];
      else
        st[node] -= a[i];
  } else {
    int mid = (l + r) >> 1;
    if (pos > mid)
      update(node << 1 | 1, mid + 1, r, pos, sign, st, a);
    else
      update(node << 1, l, mid, pos, sign, st, a);
    st[node] = max(st[node << 1], st[node << 1 | 1]);
  }
}
int get_tree(int node, int l, int r, int u, int v, vector<int>& st) {
  if (v < l || r < u) return INT_MIN;
  if (u <= l && r <= v) return st[node];
  int mid = (l + r) >> 1;
  return max(get_tree(node << 1, l, mid, u, v, st),
             get_tree(node << 1 | 1, mid + 1, r, u, v, st));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  vector<vector<int>> st((1 << k), vector<int>(4 * n + 1, 0)),
      a(n + 1, vector<int>(k));
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < k; ++j) cin >> a[i][j];
  for (int i = 0; i < (1 << k); ++i) {
    build(1, 1, n, i, st[i], a);
  }
  int q, type, pos, l, r, ans;
  cin >> q;
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> pos;
      for (int i = 0; i < k; ++i) cin >> a[pos][i];
      for (int i = 0; i < (1 << k); ++i) update(1, 1, n, pos, i, st[i], a[pos]);
    } else {
      cin >> l >> r;
      ans = 0;
      for (int i = 0; i < (1 << k); ++i)
        ans = max(ans, get_tree(1, 1, n, l, r, st[i]) +
                           get_tree(1, 1, n, l, r, st[(1 << k) - i - 1]));
      cout << ans << '\n';
    }
  }
  return 0;
}
