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 std::cin;
using std::cout;
using std::endl;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int MAXN = 200005;
int n, k, q, cnt, loc, ql, qr, xx[MAXN][6];
struct sgt {
  int maxn[33], minn[33];
} a[MAXN << 2], kk;
inline sgt mer(sgt x, sgt y) {
  for (int i = (0); i <= (cnt - 1); i++)
    x.maxn[i] = std::max(x.maxn[i], y.maxn[i]),
    x.minn[i] = std::min(x.minn[i], y.minn[i]);
  return x;
}
void build(int o, int l, int r) {
  if (l == r) {
    for (int i = (0); i <= (cnt - 1); i++) {
      for (int j = (0); j <= (k - 1); j++) {
        if ((i >> j) & 1)
          a[o].maxn[i] += xx[l][j];
        else
          a[o].maxn[i] -= xx[l][j];
      }
      a[o].minn[i] = a[o].maxn[i];
    }
    return;
  }
  build((o << 1), l, ((l + r) >> 1));
  build(((o << 1) | 1), ((l + r) >> 1) + 1, r);
  a[o] = mer(a[(o << 1)], a[((o << 1) | 1)]);
}
void upd(int o, int l, int r) {
  if (l == r) {
    a[o] = kk;
    return;
  }
  if (loc <= ((l + r) >> 1))
    upd((o << 1), l, ((l + r) >> 1));
  else
    upd(((o << 1) | 1), ((l + r) >> 1) + 1, r);
  a[o] = mer(a[(o << 1)], a[((o << 1) | 1)]);
}
sgt query(int o, int l, int r) {
  if (ql <= l && r <= qr) return a[o];
  if (((l + r) >> 1) < ql)
    return query(((o << 1) | 1), ((l + r) >> 1) + 1, r);
  else if (((l + r) >> 1) >= qr)
    return query((o << 1), l, ((l + r) >> 1));
  else
    return mer(query((o << 1), l, ((l + r) >> 1)),
               query(((o << 1) | 1), ((l + r) >> 1) + 1, r));
}
int main() {
  n = read(), k = read();
  cnt = (1 << k);
  for (int i = (1); i <= (n); i++)
    for (int j = (0); j <= (k - 1); j++) xx[i][j] = read();
  build(1, 1, n);
  q = read();
  while (q--) {
    int opt = read();
    if (opt == 1) {
      loc = read();
      for (int i = (0); i <= (k - 1); i++) xx[0][i] = read();
      for (int i = (0); i <= (cnt - 1); i++) {
        kk.maxn[i] = 0;
        for (int j = (0); j <= (k - 1); j++) {
          if ((i >> j) & 1)
            kk.maxn[i] += xx[0][j];
          else
            kk.maxn[i] -= xx[0][j];
        }
        kk.minn[i] = kk.maxn[i];
      }
      upd(1, 1, n);
    } else {
      ql = read(), qr = read();
      sgt Ans = query(1, 1, n);
      int ans = 0;
      for (int i = (0); i <= (cnt - 1); i++)
        ans = std::max(ans, Ans.maxn[i] - Ans.minn[i]);
      printf("%d\n", ans);
    }
  }
  return 0;
}
