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 maxn = 800000;
const int inf = 1000000000;
struct T {
  int Mx[maxn], Mn[maxn];
  void update(int Now, int L, int R, int pos, int val) {
    if (L == R) {
      Mn[Now] = Mx[Now] = val;
      return;
    }
    int Mid = (L + R) >> 1;
    if (pos <= Mid)
      update(Now << 1, L, Mid, pos, val);
    else
      update(Now << 1 | 1, Mid + 1, R, pos, val);
    Mx[Now] = max(Mx[Now << 1], Mx[Now << 1 | 1]);
    Mn[Now] = min(Mn[Now << 1], Mn[Now << 1 | 1]);
  }
  pair<int, int> query(int Now, int L, int R, int l, int r) {
    if (l <= L && r >= R) return make_pair(Mx[Now], Mn[Now]);
    int Mid = (L + R) >> 1;
    int mx = -inf, mn = inf;
    pair<int, int> tmp;
    if (l <= Mid) {
      tmp = query(Now << 1, L, Mid, l, r);
      mx = max(mx, tmp.first);
      mn = min(mn, tmp.second);
    }
    if (r > Mid) {
      tmp = query(Now << 1 | 1, Mid + 1, R, l, r);
      mx = max(mx, tmp.first);
      mn = min(mn, tmp.second);
    }
    return make_pair(mx, mn);
  }
} t[16];
int main() {
  int N, K, M, opt, L, R, p, x[6], KK;
  scanf("%d%d", &N, &K);
  KK = (1 << (K - 1)) - 1;
  for (int i = 0; i <= KK; i++) {
    for (int j = 0; j <= N * 4; j++) t[i].Mn[j] = inf, t[i].Mx[j] = -inf;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j <= K - 1; j++) scanf("%d", &x[j]);
    for (int j = 0; j <= KK; j++) {
      int tmp = 0;
      for (int k = 0; k <= K - 1; k++)
        if (j & (1 << k))
          tmp += x[k];
        else
          tmp -= x[k];
      t[j].update(1, 1, N, i, tmp);
    }
  }
  scanf("%d", &M);
  for (int i = 1; i <= M; i++) {
    scanf("%d", &opt);
    if (opt == 2) {
      scanf("%d%d", &L, &R);
      int ans = -inf;
      pair<int, int> tmp;
      for (int j = 0; j <= KK; j++) {
        pair<int, int> tmp = t[j].query(1, 1, N, L, R);
        ans = max(ans, tmp.first - tmp.second);
      }
      printf("%d\n", ans);
    } else {
      scanf("%d", &p);
      for (int j = 0; j <= K - 1; j++) scanf("%d", &x[j]);
      for (int j = 0; j <= KK; j++) {
        int tmp = 0;
        for (int k = 0; k <= K - 1; k++)
          if (j & (1 << k))
            tmp += x[k];
          else
            tmp -= x[k];
        t[j].update(1, 1, N, p, tmp);
      }
    }
  }
  return 0;
}
