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;
int n, k, q, x, y, t, full;
int S[200005 * 4][32];
vector<int> a[200005];
int mul[2] = {-1, 1};
void merge(int node, int l, int r) {
  for (int mask = 0; mask <= full; mask++) {
    S[node][mask] = max(S[l][mask], S[r][mask]);
  }
}
void create(int node, int bas) {
  for (int mask = 0; mask <= full; mask++) {
    S[node][mask] = 0;
    for (int i = 0; i < k; i++) {
      S[node][mask] += a[bas][i] * mul[((1 << (i)) & mask) > 0];
    }
  }
}
int get(int node, int bas, int son, int x, int y, int mask) {
  if (bas >= x && son <= y) return S[node][mask];
  if (bas > y || son < x) return -2000000000;
  return max(get(node * 2, bas, ((bas + son) >> 1), x, y, mask),
             get(node * 2 + 1, ((bas + son) >> 1) + 1, son, x, y, mask));
}
void up(int node, int bas, int son, int x) {
  if (bas > x || son < x) return;
  if (bas == x && son == x) {
    create(node, bas);
    return;
  }
  up(node * 2, bas, ((bas + son) >> 1), x);
  up(node * 2 + 1, ((bas + son) >> 1) + 1, son, x);
  merge(node, node * 2, node * 2 + 1);
}
void build(int node, int bas, int son) {
  if (bas == son) {
    create(node, bas);
    return;
  }
  build(node * 2, bas, ((bas + son) >> 1));
  build(node * 2 + 1, ((bas + son) >> 1) + 1, son);
  merge(node, node * 2, node * 2 + 1);
}
int main() {
  scanf("%d %d", &n, &k);
  full = (1 << (k)) - 1;
  for (int i = 1; i <= n; i++) {
    a[i].resize(k);
    for (int j = 0; j < k; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d", &t);
      for (int i = 0; i < k; i++) scanf("%d", &a[t][i]);
      up(1, 1, n, t);
    } else {
      scanf("%d %d", &x, &y);
      int ans = 0;
      for (int mask = 0; mask <= full / 2; mask++) {
        ans = max(ans,
                  get(1, 1, n, x, y, mask) + get(1, 1, n, x, y, full ^ mask));
      }
      printf("%d\n", ans);
    }
  }
}
