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;
using ll = long long;
using ii = pair<int, int>;
const int N = 200005;
class SegTree {
  vector<int> st;
  vector<int> st2;
  int n;
  void upd(int p, int nodeL, int nodeR, int queryL, int queryR, int v) {
    if (queryR < nodeL or queryL > nodeR) return;
    if (queryL <= nodeL and queryR >= nodeR) {
      st[p] = v;
      st2[p] = v;
      return;
    }
    int mid = (nodeL + nodeR) / 2;
    upd(2 * p, nodeL, mid, queryL, queryR, v);
    upd(2 * p + 1, mid + 1, nodeR, queryL, queryR, v);
    st[p] = max(st[2 * p], st[2 * p + 1]);
    st2[p] = min(st2[2 * p], st2[2 * p + 1]);
  }
  int qry(int p, int nodeL, int nodeR, int queryL, int queryR) {
    if (queryR < nodeL or queryL > nodeR) return -1e9;
    if (queryL <= nodeL and queryR >= nodeR) return st[p];
    int mid = (nodeL + nodeR) / 2;
    return max(qry(2 * p, nodeL, mid, queryL, queryR),
               qry(2 * p + 1, mid + 1, nodeR, queryL, queryR));
  }
  int qry2(int p, int nodeL, int nodeR, int queryL, int queryR) {
    if (queryR < nodeL or queryL > nodeR) return 1e9;
    if (queryL <= nodeL and queryR >= nodeR) return st2[p];
    int mid = (nodeL + nodeR) / 2;
    return min(qry2(2 * p, nodeL, mid, queryL, queryR),
               qry2(2 * p + 1, mid + 1, nodeR, queryL, queryR));
  }

 public:
  SegTree(int sz) {
    n = sz;
    st.assign(5 * (n + 1), -1e9);
    st2.assign(5 * (n + 1), 1e9);
  }
  int qry(int i, int j) { return qry(1, 1, n, i, j); }
  int qry2(int i, int j) { return qry2(1, 1, n, i, j); }
  void upd(int i, int j, int v) { upd(1, 1, n, i, j, v); }
};
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  vector<SegTree> st(40, SegTree(n));
  int a[N][6];
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &a[i][j]);
    }
    for (int j = 0; j < (1 << k); j++) {
      int sum = 0;
      for (int l = 0; l < k; l++) {
        if (j & (1 << l))
          sum += a[i][l];
        else
          sum -= a[i][l];
      }
      st[j].upd(i, i, sum);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int id;
      scanf("%d", &id);
      for (int j = 0; j < k; j++) {
        scanf("%d", &a[id][j]);
      }
      for (int j = 0; j < (1 << k); j++) {
        int sum = 0;
        for (int l = 0; l < k; l++) {
          if (j & (1 << l))
            sum += a[id][l];
          else
            sum -= a[id][l];
        }
        st[j].upd(id, id, sum);
      }
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      ll ans = -1e9;
      for (int j = 0; j < (1 << k); j++) {
        ans = max(ans, 1LL * st[j].qry(l, r) - st[j].qry2(l, r));
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
