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;
template <typename T1, typename T2>
bool mini(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
bool maxi(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
const int N = 2e5 + 5;
const int oo = 1e9;
long long mx[1 << 5];
long long st[N << 2][1 << 5];
int a[N][5];
int n, k, q, timer;
int bit(int x) { return (1 << x); }
void update(int id, int l, int r, int pos, int a[]) {
  if (l > pos || r < pos) return;
  if (l == r) {
    for (int mask = 0; mask < bit(k); mask++) {
      st[id][mask] = 0;
      int x = mask;
      for (int i = 0; i < k; i++) {
        if (x & 1)
          st[id][mask] += a[i];
        else
          st[id][mask] -= a[i];
        x >>= 1;
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  update(id << 1, l, mid, pos, a);
  update(id << 1 | 1, mid + 1, r, pos, a);
  for (int mask = 0; mask < bit(k); mask++)
    st[id][mask] = max(st[id << 1][mask], st[id << 1 | 1][mask]);
}
void getmax(int id, int l, int r, int u, int v) {
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    for (int mask = 0; mask < bit(k); mask++) maxi(mx[mask], st[id][mask]);
    return;
  }
  int mid = (l + r) >> 1;
  getmax(id << 1, l, mid, u, v);
  getmax(id << 1 | 1, mid + 1, r, u, v);
}
void getmax(int l, int r) {
  for (int mask = 0; mask < bit(k); mask++) mx[mask] = -(long long)oo * oo;
  getmax(1, 1, n, l, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++) update(1, 1, n, i, a[i]);
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int v;
      cin >> v;
      for (int i = 0; i < k; i++) cin >> a[v][i];
      update(1, 1, n, v, a[v]);
    } else {
      int l, r;
      cin >> l >> r;
      getmax(l, r);
      long long ans = 0;
      for (int mask = 0; mask < bit(k); mask++)
        maxi(ans, mx[mask] + mx[mask ^ (bit(k) - 1)]);
      cout << ans << "\n";
    }
  }
  return 0;
}
