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 = (int)2e5 + 5;
const int M = 1 << 18;
const int K = 5;
const int L = 1 << K;
const int INF = (int)1e9;
int n, k;
int tree[L][2 * M];
int get(int v, int tl, int tr, int mask, int l, int r) {
  if (tr <= l || r <= tl) return -INF;
  if (l <= tl && tr <= r) return tree[mask][v];
  int tm = (tl + tr) / 2;
  int ansL = get(2 * v, tl, tm, mask, l, r);
  int ansR = get(2 * v + 1, tm, tr, mask, l, r);
  return max(ansL, ansR);
}
int get(int l, int r, int mask) {
  int res = get(1, 0, M, mask, l, r);
  return res;
}
void update(int v, int tl, int tr, int pos, int mask, int new_val) {
  if (pos < tl || tr <= pos) return;
  if (tl + 1 == tr) {
    tree[mask][v] = new_val;
    return;
  }
  int tm = (tl + tr) / 2;
  update(2 * v, tl, tm, pos, mask, new_val);
  update(2 * v + 1, tm, tr, pos, mask, new_val);
  tree[mask][v] = max(tree[mask][2 * v], tree[mask][2 * v + 1]);
}
int a[K];
void upd(int pos) {
  for (int i = 0; i < k; ++i) scanf("%d", a + i);
  for (int mask = 0; mask < (1 << k); ++mask) {
    int val = 0;
    for (int i = 0; i < k; ++i) {
      if ((mask >> i) & 1)
        val += a[i];
      else
        val -= a[i];
    }
    update(1, 0, M, pos, mask, val);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int mask = 0; mask < (1 << k); ++mask) {
    for (int i = 1; i < 2 * M; ++i) tree[mask][i] = -INF;
  }
  for (int i = 0; i < n; ++i) upd(i);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int pos;
      scanf("%d", &pos);
      --pos;
      upd(pos);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      --l;
      --r;
      int ans = 0;
      for (int mask = 0; mask < (1 << k); mask += 2) {
        ans = max(ans,
                  get(l, r + 1, mask) + get(l, r + 1, mask ^ ((1 << k) - 1)));
      }
      printf("%d\n", ans);
    }
  }
}
