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>
const int N = 200005, M = 5;
using namespace std;
int n, k, q;
struct T {
  vector<int> mn, mx;
  T() {
    mn = vector<int>(4 * N, 0);
    mx = vector<int>(4 * N, 0);
  }
  void upd(int i, int l, int r, int pos, int vl) {
    if (l == r) {
      mn[i] = mx[i] = vl;
      return;
    }
    int m = (l + r) / 2;
    if (pos <= m)
      upd(i * 2, l, m, pos, vl);
    else
      upd(i * 2 + 1, m + 1, r, pos, vl);
    mn[i] = min(mn[i * 2], mn[i * 2 + 1]);
    mx[i] = max(mx[i * 2 + 1], mx[i * 2 + 1]);
  }
  pair<int, int> get(int i, int l, int r, int se, int en) {
    if (se <= l && r <= en) return {mn[i], mx[i]};
    if (l > en || r < se) return {INT_MAX, INT_MIN};
    int m = (l + r) / 2;
    pair<int, int> a = get(i * 2, l, m, se, en);
    pair<int, int> b = get(i * 2 + 1, m + 1, r, se, en);
    return {min(a.first, b.first), max(a.second, b.second)};
  }
} dp[32];
array<int, M> ar[N];
int vl(int msk, array<int, M>& v) {
  int ans = 0;
  for (int i = 0; i < k; i++) ans += ((msk & (1 << i)) ? 1 : -1) * v[i];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) cin >> ar[i][j];
    for (int j = 0; j < (1 << k); j++) dp[j].upd(1, 1, n, i, vl(j, ar[i]));
  }
  int mx = (1 << k) - 1;
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int i;
      cin >> i;
      for (int j = 0; j < k; j++) cin >> ar[i][j];
      for (int j = 0; j < (1 << k); j++) dp[j].upd(1, 1, n, i, vl(j, ar[i]));
    } else {
      int l, r;
      cin >> l >> r;
      int ans = 0;
      for (int j = 0; j < (1 << k); j++) {
        auto e1 = dp[j].get(1, 1, n, l, r);
        auto e2 = dp[j xor mx].get(1, 1, n, l, r);
        ans = max(ans, abs(e1.first + e2.first));
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
