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 + 10;
struct node {
  long long v[1 << 5];
  node() { memset(v, 0, sizeof v); }
} t[N << 2];
int k;
long long a[N][5];
node merge(node a, node b) {
  for (int i = 0; i < (1 << k); i++) a.v[i] = max(a.v[i], b.v[i]);
  return a;
}
void build(int l, int r, int pos) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      for (int j = 0; j < k; j++)
        if (i >> j & 1)
          t[pos].v[i] += a[l][j];
        else
          t[pos].v[i] -= a[l][j];
    }
    return;
  }
  int m = l + r >> 1;
  build(l, m, pos << 1);
  build(m + 1, r, pos << 1 | 1);
  t[pos] = merge(t[pos << 1], t[pos << 1 | 1]);
}
long long temp[5];
void update(int x, int l, int r, int pos) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      t[pos].v[i] = 0;
      for (int j = 0; j < k; j++)
        if (i >> j & 1)
          t[pos].v[i] += temp[j];
        else
          t[pos].v[i] -= temp[j];
    }
    return;
  }
  int m = l + r >> 1;
  if (x <= m)
    update(x, l, m, pos << 1);
  else
    update(x, m + 1, r, pos << 1 | 1);
  t[pos] = merge(t[pos << 1], t[pos << 1 | 1]);
}
node qmax(int L, int R, int l, int r, int pos) {
  if (L <= l && r <= R) {
    return t[pos];
  }
  node ans;
  for (int i = 0; i < (1 << k); i++) ans.v[i] = -1e18;
  int m = l + r >> 1;
  if (L <= m) ans = merge(ans, qmax(L, R, l, m, pos << 1));
  if (R > m) ans = merge(ans, qmax(L, R, m + 1, r, pos << 1 | 1));
  return ans;
}
int n;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < k; j++) scanf("%lld", &a[i][j]);
  build(1, n, 1);
  int m;
  cin >> m;
  int op, l, r, x;
  while (m--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &x);
      for (int i = 0; i < k; i++) scanf("%lld", &temp[i]);
      update(x, 1, n, 1);
    } else {
      scanf("%d%d", &l, &r);
      node x = qmax(l, r, 1, n, 1);
      long long ans = 0;
      for (int i = 0; i < (1 << k); i++)
        ans = max(ans, x.v[i] + x.v[(1 << k) - 1 - i]);
      printf("%lld\n", ans);
    }
  }
}
