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 + 5;
const long long mod = 1e9 + 7;
int t[N << 2][33];
int a[N][6];
int n, k;
void build(int rt, int l, int r) {
  for (int i = 0; i < (1 << k); i++) t[rt][i] = -0x3fffffff;
  if (l == r) {
    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[rt][i] = cur;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  for (int i = 0; i < (1 << k); i++)
    t[rt][i] = max(t[rt << 1][i], t[rt << 1 | 1][i]);
}
void update(int rt, int l, int r, int pos) {
  if (l == r && l == pos) {
    for (int i = 0; i < (1 << k); i++) {
      int cur = 0;
      for (int j = 0; j < k; j++) {
        if (i & (1 << j))
          cur += a[pos][j];
        else
          cur -= a[pos][j];
      }
      t[rt][i] = cur;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update(rt << 1, l, mid, pos);
  else
    update(rt << 1 | 1, mid + 1, r, pos);
  for (int i = 0; i < (1 << k); i++)
    t[rt][i] = max(t[rt << 1][i], t[rt << 1 | 1][i]);
}
int query(int rt, int l, int r, int L, int R, int sta) {
  if (L <= l && r <= R) return t[rt][sta];
  int mid = (l + r) >> 1;
  int ans = -0x3fffffff;
  if (L <= mid) ans = max(ans, query(rt << 1, l, mid, L, R, sta));
  if (R > mid) ans = max(ans, query(rt << 1 | 1, mid + 1, r, L, R, sta));
  return ans;
}
void solve() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= k - 1; j++) scanf("%d", &a[i][j]);
  build(1, 1, n);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos;
      scanf("%d", &pos);
      for (int j = 0; j <= k - 1; j++) scanf("%d", &a[pos][j]);
      update(1, 1, n, pos);
    } else {
      int ans = 0;
      int l, r;
      scanf("%d %d", &l, &r);
      for (int i = 0; i < (1 << k); i++) {
        ans = max(ans, query(1, 1, n, l, r, i) +
                           query(1, 1, n, l, r, (1 << k) - 1 - i));
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  solve();
  return 0;
}
