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;
const int N = 2e5 + 5;
int a[N][6];
int t1, t2;
inline void kma(int a, int b, int &c) { c = a > b ? a : b; }
inline void kmi(int a, int b, int &c) { c = a > b ? b : a; }
struct SegT {
  int ma[N * 4], mi[N * 4];
  inline void up(int i) {
    kma(ma[i * 2], ma[i * 2 + 1], ma[i]);
    kmi(mi[i * 2], mi[i * 2 + 1], mi[i]);
  }
  void ins(int i, int l, int r, int pos, int v) {
    if (l == r) {
      ma[i] = mi[i] = v;
      return;
    }
    int mid = (l + r) / 2;
    pos <= mid ? ins(i * 2, l, mid, pos, v)
               : ins(i * 2 + 1, mid + 1, r, pos, v);
    up(i);
  }
  int ans1, ans2;
  void query(int i, int l, int r, int L, int R) {
    if (L <= l && r <= R) {
      kma(ans1, ma[i], ans1);
      kmi(ans2, mi[i], ans2);
      return;
    }
    int mid = (l + r) / 2;
    if (L <= mid) query(i * 2, l, mid, L, R);
    if (R > mid) query(i * 2 + 1, mid + 1, r, L, R);
  }
} sg[33];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int le = 1 << k;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) scanf("%d", &a[i][j]);
    for (int s = 0; s < le; ++s) {
      int tmp = 0;
      for (int j = 0; j < k; ++j)
        if (s & (1 << j))
          tmp += a[i][j];
        else
          tmp -= a[i][j];
      sg[s].ins(1, 1, n, i, tmp);
    }
  }
  int t[5], i;
  int q;
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &i);
      for (int j = 0; j < k; ++j) scanf("%d", &t[j]);
      for (int s = 0; s < le; ++s) {
        int tmp = 0;
        for (int j = 0; j < k; ++j)
          if (s & (1 << j))
            tmp += t[j];
          else
            tmp -= t[j];
        sg[s].ins(1, 1, n, i, tmp);
      }
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int s = 0; s < le; ++s) {
        sg[s].ans1 = -1e9, sg[s].ans2 = 1e9;
        sg[s].query(1, 1, n, l, r);
        kma(ans, sg[s].ans1 - sg[s].ans2, ans);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
