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 = 2e5 + 7, K = 33, inf = -1e9;
struct tree {
  int mx[N << 2];
  tree() {
    for (int i = 0; i < (N << 2); ++i) mx[i] = inf;
  }
  void update(int rt) { mx[rt] = max(mx[(rt << 1)], mx[(rt << 1 | 1)]); }
  void modify(int rt, int l, int r, int pos, int val) {
    if (l == r) {
      mx[rt] = val;
      return;
    }
    int mid = l + r >> 1;
    if (pos <= mid)
      modify((rt << 1), l, mid, pos, val);
    else
      modify((rt << 1 | 1), mid + 1, r, pos, val);
    update(rt);
  }
  int query(int rt, int l, int r, int a, int b) {
    if (a <= l && r <= b) return mx[rt];
    int res = inf, mid = l + r >> 1;
    if (a <= mid) res = query((rt << 1), l, mid, a, b);
    if (b > mid) res = max(res, query((rt << 1 | 1), mid + 1, r, a, b));
    return res;
  }
} t[K];
int n, k, i, j, a[N][7], ans[K], op, res = inf, m, l, r;
void modify(int id) {
  for (int i = 0, j, cnt; i < (1 << k); ++i) {
    for (j = cnt = 0; j < k; ++j) cnt += (i >> j & 1) ? a[id][j] : -a[id][j];
    t[i].modify(1, 1, n, id, cnt);
  }
}
int main() {
  for (scanf("%d%d", &n, &k), i = 1; i <= n; modify(i++))
    for (j = 0; j < k; ++j) scanf("%d", &a[i][j]);
  for (scanf("%d", &m); m--;) {
    scanf("%d", &op);
    if (op == 1) {
      for (scanf("%d", &i), j = 0; j < k; ++j) scanf("%d", &a[i][j]);
      modify(i);
    } else {
      for (scanf("%d%d", &l, &r), i = 0; i < (1 << k); ++i)
        ans[i] = t[i].query(1, 1, n, l, r);
      for (res = inf, i = 0; i < (1 << k); ++i)
        res = max(res, ans[i] + ans[(1 << k) - 1 - i]);
      printf("%d\n", res);
    }
  }
}
