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;
int a[400005][5];
int c[2] = {-1, 1};
int mx[800005][1 << 5];
int calc(int* x, int s) {
  int ret = 0;
  for (int i = 0; i < k; i++) {
    ret += c[s >> i & 1] * x[i];
  }
  return ret;
}
void merge(int o) {
  for (int i = 0; i < (1 << k); i++) {
    mx[o][i] = max(mx[o << 1][i], mx[o << 1 | 1][i]);
  }
}
void build(int o, int l, int r) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      mx[o][i] = calc(a[l], i);
    }
  } else {
    int mid = (l + r) / 2;
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    merge(o);
  }
}
void cg(int o, int l, int r, int pos, int* x) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      mx[o][i] = calc(x, i);
    }
  } else {
    int mid = (l + r) / 2;
    if (pos <= mid)
      cg(o << 1, l, mid, pos, x);
    else
      cg(o << 1 | 1, mid + 1, r, pos, x);
    merge(o);
  }
}
int ask(int o, int l, int r, int L, int R, int k) {
  if (l == L && r == R) {
    return mx[o][k];
  } else {
    int mid = (l + r) / 2;
    if (R <= mid)
      return ask(o << 1, l, mid, L, R, k);
    else if (L > mid)
      return ask(o << 1 | 1, mid + 1, r, L, R, k);
    else
      return max(ask(o << 1, l, mid, L, mid, k),
                 ask(o << 1 | 1, mid + 1, r, mid + 1, R, k));
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) scanf("%d", &a[i][j]);
  int q;
  scanf("%d", &q);
  build(1, 1, n);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int x;
      int tmp[5];
      scanf("%d", &x);
      for (int i = 0; i < k; i++) scanf("%d", &tmp[i]);
      cg(1, 1, n, x, tmp);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int mx = 0;
      for (int i = 0; i < (1 << k); i++) {
        mx = max(
            mx, ask(1, 1, n, l, r, i) + ask(1, 1, n, l, r, i ^ ((1 << k) - 1)));
      }
      printf("%d\n", mx);
    }
  }
  return 0;
}
