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 N = 200043;
const int INF = int(1e9);
int T[4 * N][32];
int A[4 * N][5];
int n, k;
void build(int v, int l, int r) {
  if (l == r - 1) {
    for (int i = 0; i < (1 << k); i++) {
      int cur = 0;
      for (int j = 0; j < k; j++)
        if (i & (1 << j))
          cur += A[l][j];
        else
          cur -= A[l][j];
      T[v][i] = cur;
    }
  } else {
    int m = (l + r) / 2;
    build(v * 2 + 1, l, m);
    build(v * 2 + 2, m, r);
    for (int i = 0; i < (1 << k); i++)
      T[v][i] = min(T[v * 2 + 1][i], T[v * 2 + 2][i]);
  }
}
int query(int v, int l, int r, int L, int R, int x) {
  if (L >= R) return INF;
  if (L == l && R == r) return T[v][x];
  int m = (l + r) / 2;
  return min(query(v * 2 + 1, l, m, L, min(R, m), x),
             query(v * 2 + 2, m, r, max(L, m), R, x));
}
void upd(int v, int l, int r, int z) {
  if (l == r - 1) {
    for (int i = 0; i < (1 << k); i++) {
      int cur = 0;
      for (int j = 0; j < k; j++)
        if (i & (1 << j))
          cur += A[l][j];
        else
          cur -= A[l][j];
      T[v][i] = cur;
    }
  } else {
    int m = (l + r) / 2;
    if (z < m)
      upd(v * 2 + 1, l, m, z);
    else
      upd(v * 2 + 2, m, r, z);
    for (int i = 0; i < (1 << k); i++)
      T[v][i] = min(T[v * 2 + 1][i], T[v * 2 + 2][i]);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &A[i][j]);
    }
  }
  int q;
  scanf("%d", &q);
  build(0, 0, n);
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int z;
      scanf("%d", &z);
      --z;
      for (int j = 0; j < k; j++) scanf("%d", &A[z][j]);
      upd(0, 0, n, z);
    } else {
      int ans = 0;
      int l, r;
      scanf("%d %d", &l, &r);
      --l;
      for (int i = 0; i < (1 << (k - 1)); i++) {
        int x = query(0, 0, n, l, r, i);
        int y = query(0, 0, n, l, r, ((1 << k) - 1) ^ i);
        ans = max(ans, abs(x + y));
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
