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 Nmax = 2e5 + 11;
const int Bmax = (1 << 5) + 2;
const int Kmax = 6;
const int inf = 1e9;
int n, k, ver[Bmax], q;
pair<int, int> IT[Bmax][4 * Nmax];
int getbit(int a, int k) { return (a >> (k - 1)) & 1; }
struct point {
  int b[Kmax];
  int val(int mask) {
    int res = 0;
    for (int i = (int)1; i <= (int)k; ++i) {
      if (getbit(mask, i) == 0)
        res += b[i];
      else
        res -= b[i];
    }
    return res;
  }
} a[Nmax];
void update(int id, int l, int r, int L, int R, int mask, point a) {
  if (l > R || r < L) return;
  if (L <= l && r <= R) {
    IT[mask][id].first = IT[mask][id].second = a.val(mask);
    return;
  }
  int mid = (l + r) >> 1;
  update(id * 2, l, mid, L, R, mask, a);
  update(id * 2 + 1, mid + 1, r, L, R, mask, a);
  IT[mask][id].first = max(IT[mask][id * 2].first, IT[mask][id * 2 + 1].first);
  IT[mask][id].second =
      min(IT[mask][id * 2].second, IT[mask][id * 2 + 1].second);
}
pair<int, int> get(int id, int l, int r, int L, int R, int mask) {
  if (l > R || r < L) return pair<int, int>(-inf, inf);
  if (L <= l && r <= R) {
    return IT[mask][id];
  }
  int mid = (l + r) >> 1;
  pair<int, int> res1 = get(id * 2, l, mid, L, R, mask);
  pair<int, int> res2 = get(id * 2 + 1, mid + 1, r, L, R, mask);
  return pair<int, int>(max(res1.first, res2.first),
                        min(res1.second, res2.second));
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = (int)0; i < (int)1 << k; ++i) {
    int tmp = i;
    for (int j = (int)1; j <= (int)k; ++j) {
      if (getbit(i, j))
        tmp -= (1 << (j - 1));
      else
        tmp += (1 << (j - 1));
    }
    ver[i] = tmp;
  }
  for (int i = (int)1; i <= (int)n; ++i)
    for (int j = (int)1; j <= (int)k; ++j) {
      scanf("%d", &a[i].b[j]);
    }
  for (int i = (int)1; i <= (int)n; ++i)
    for (int l = (int)0; l < (int)1 << k; ++l) {
      update(1, 1, n, i, i, l, a[i]);
    }
  scanf("%d", &q);
  for (int i = (int)1; i <= (int)q; ++i) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int pos;
      scanf("%d", &pos);
      point tmp;
      for (int j = (int)1; j <= (int)k; ++j) scanf("%d", &tmp.b[j]);
      for (int j = (int)0; j < (int)1 << k; ++j)
        update(1, 1, n, pos, pos, j, tmp);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      int res = 0;
      for (int j = (int)0; j < (int)1 << k; ++j) {
        pair<int, int> tmp = get(1, 1, n, l, r, j);
        res = max(res, tmp.first - tmp.second);
      }
      cout << res << '\n';
    }
  }
}
