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 double pi = 3.14159265358979323846;
#pragma GCC optimize("-O2")
const int LM = 2e5, inf = 1e9;
int q, n, k, x, l, r, a[6], maxseg[33][4 * LM + 1], minseg[33][4 * LM + 1],
    siz = 1;
void upd(int x[], int pos) {
  for (int i = 0; i <= (1 << k) - 1; i++) {
    int t = i, an = 0;
    for (int j = 1; j <= k; j++) {
      if ((t & 1) == 1) {
        an += x[j];
      } else {
        an -= x[j];
      }
      t /= 2;
    }
    int ps = pos + siz - 1;
    minseg[i][ps] = an;
    maxseg[i][ps] = an;
    ps /= 2;
    while (ps > 0) {
      minseg[i][ps] = min(minseg[i][2 * ps], minseg[i][2 * ps + 1]);
      maxseg[i][ps] = max(maxseg[i][2 * ps], maxseg[i][2 * ps + 1]);
      ps /= 2;
    }
  }
}
int query(int l, int r) {
  int ret = -inf;
  for (int i = 0; i <= (1 << k) - 1; i++) {
    int a1 = -inf, a2 = inf;
    int l1 = siz + l - 1, r1 = siz + r - 1;
    int l2 = siz + l - 1, r2 = siz + r - 1;
    while (l1 <= r1) {
      if (l1 % 2 == 1) a1 = max(a1, maxseg[i][l1++]);
      if (r1 % 2 == 0) a1 = max(a1, maxseg[i][r1--]);
      l1 /= 2;
      r1 /= 2;
    }
    while (l2 <= r2) {
      if (l2 % 2 == 1) a2 = min(a2, minseg[i][l2++]);
      if (r2 % 2 == 0) a2 = min(a2, minseg[i][r2--]);
      l2 /= 2;
      r2 /= 2;
    }
    ret = max(ret, a1 - a2);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> k;
  while (siz < n) siz <<= 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      cin >> a[j];
    }
    upd(a, i);
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> x;
    if (x == 1) {
      int pos;
      cin >> pos;
      for (int j = 1; j <= k; j++) {
        cin >> a[j];
      }
      upd(a, pos);
    } else {
      cin >> l >> r;
      cout << query(l, r) << endl;
    }
  }
}
