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>
int n, k, a[10], m;
struct P {
  long long a[800010];
};
P t[40];
int Max(long long x, long long y) { return x > y ? x : y; }
void mdy(int u, int l, int r, int x, long long w, int id) {
  if (l >= r) {
    t[id].a[u] = w;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    mdy(u * 2, l, mid, x, w, id);
  else
    mdy(u * 2 + 1, mid + 1, r, x, w, id);
  t[id].a[u] = Max(t[id].a[u * 2], t[id].a[u * 2 + 1]);
}
void update(int x) {
  for (int i = 1; i <= k; i++) scanf("%d", &a[i]);
  long long sum;
  for (int i = 0; i < (1 << k); i++) {
    sum = 0;
    for (int j = 0; j < k; j++)
      sum += (long long)a[j + 1] * ((i & (1 << j)) ? -1ll : 1ll);
    mdy(1, 1, n, x, sum, i);
  }
}
long long ask(int u, int l, int r, int L, int R, int id) {
  if (l == L && r == R) return t[id].a[u];
  int mid = (l + r) >> 1;
  if (R <= mid) return ask(u * 2, l, mid, L, R, id);
  if (L > mid) return ask(u * 2 + 1, mid + 1, r, L, R, id);
  return Max(ask(u * 2, l, mid, L, mid, id),
             ask(u * 2 + 1, mid + 1, r, mid + 1, R, id));
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) update(i);
  scanf("%d", &m);
  for (; m; --m) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      int x;
      scanf("%d", &x), update(x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      long long ans = -0x3f3f3f3f;
      for (int i = 0; i < (1 << k); i++) {
        if (i > ((1 << k) - 1 - i)) continue;
        ans = Max(ans, ask(1, 1, n, l, r, i) +
                           ask(1, 1, n, l, r, ((1 << k) - 1 - i)));
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
