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 dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
int a[100005 * 2][6], val[6];
int sum[33][100005 * 8], n, m;
void build(int l, int r, int k) {
  if (l == r) {
    for (int i = 0; i < (1 << m); i++) {
      int tot = 0;
      for (int j = 0; j < m; j++) {
        if ((bool)(i & 1 << j))
          tot += a[l][j];
        else
          tot -= a[l][j];
      }
      sum[i][k] = tot;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, k * 2);
  build(mid + 1, r, k * 2 + 1);
  for (int i = 0; i < (1 << m); i++) {
    sum[i][k] = min(sum[i][k * 2], sum[i][k * 2 + 1]);
  }
}
void update(int l, int r, int k, int L) {
  if (l == r) {
    for (int i = 0; i < (1 << m); i++) {
      long long tot = 0;
      for (int j = 0; j < m; j++) {
        if ((bool)(i & 1 << j))
          tot += val[j];
        else
          tot -= val[j];
      }
      sum[i][k] = tot;
    }
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= L) {
    update(l, mid, k * 2, L);
  } else {
    update(mid + 1, r, k * 2 + 1, L);
  }
  for (int i = 0; i < (1 << m); i++) {
    sum[i][k] = min(sum[i][k * 2], sum[i][k * 2 + 1]);
  }
}
long long query(int mask, int l, int r, int k, int L, int R) {
  if (l > R || r < L) return 2134567891;
  ;
  if (l >= L && r <= R) return sum[mask][k];
  int mid = (l + r) / 2;
  return min(query(mask, l, mid, k * 2, L, R),
             query(mask, mid + 1, r, k * 2 + 1, L, R));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) scanf("%d", &a[i + 1][j]);
  }
  build(1, n, 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int typ;
    scanf("%d", &typ);
    if (typ == 1) {
      int pos;
      scanf("%d", &pos);
      for (int i = 0; i < m; i++) scanf("%d", &val[i]);
      update(1, n, 1, pos);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      int mx = 0;
      int mask = (1 << m) - 1;
      for (int i = 0; i < (1 << (m - 1)); i++) {
        int ret = query(i, 1, n, 1, l, r);
        int rett = query(mask ^ i, 1, n, 1, l, r);
        mx = max(mx, -rett - ret);
      }
      printf("%d\n", (mx));
    }
  }
  return 0;
}
