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>
#pragma GCC optimize(2)
using namespace std;
const int N = 2e5 + 10;
struct zxq {
  int seg[N << 2];
  inline void pushup(int rt) { seg[rt] = max(seg[rt << 1], seg[rt << 1 | 1]); }
  void motify(int rt, int l, int r, int k, int d) {
    if (l == r) {
      seg[rt] = d;
      return;
    }
    int mid = l + r >> 1;
    if (mid >= k)
      motify(rt << 1, l, mid, k, d);
    else
      motify(rt << 1 | 1, mid + 1, r, k, d);
    pushup(rt);
  }
  int query(int rt, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return seg[rt];
    int mid = l + r >> 1;
    int res = 1 << 31;
    if (mid >= ql) res = max(res, query(rt << 1, l, mid, ql, qr));
    if (mid + 1 <= qr) res = max(res, query(rt << 1 | 1, mid + 1, r, ql, qr));
    return res;
  }
} tr[32];
int n, q, k, a[N][6], p[2], nc[6], idx;
int main() {
  scanf("%d%d", &n, &k);
  register int i, j, s;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= k; ++j) scanf("%d", &a[i][j]);
  p[0] = -1;
  p[1] = 1;
  for (i = 1; i <= n; ++i) {
    for (j = 0; j <= 31; ++j) {
      int tmp = j, sum = 0;
      for (s = 1; s <= 5; ++s) sum += p[tmp & 1] * a[i][s], tmp >>= 1;
      tr[j].motify(1, 1, n, i, sum);
    }
  }
  scanf("%d", &q);
  while (q--) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d", &idx);
      for (i = 1; i <= k; ++i) scanf("%d", &nc[i]);
      for (j = 0; j <= 31; ++j) {
        int tmp = j, sum = 0;
        for (s = 1; s <= 5; ++s) sum += p[tmp & 1] * nc[s], tmp >>= 1;
        tr[j].motify(1, 1, n, idx, sum);
      }
    } else {
      int L, R;
      scanf("%d%d", &L, &R);
      int ans = 0;
      for (i = 0; i <= 16; ++i) {
        ans = max(ans,
                  tr[i].query(1, 1, n, L, R) + tr[31 - i].query(1, 1, n, L, R));
      }
      printf("%d\n", ans);
    }
  }
}
