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;
int a[10];
struct st {
  pair<int, int> st[800005];
  void update(int n, int s, int e, int idx, int v) {
    if (s == e)
      st[n] = pair<int, int>(v, v);
    else {
      int mid = (s + e) / 2;
      if (idx <= mid)
        update(n + n, s, mid, idx, v);
      else
        update(n + n + 1, mid + 1, e, idx, v);
      st[n] = pair<int, int>(max(st[n + n].first, st[n + n + 1].first),
                             min(st[n + n].second, st[n + n + 1].second));
    }
  }
  pair<int, int> query(int n, int s, int e, int l, int r) {
    if (s > r || l > e) return pair<int, int>(-2e9, 2e9);
    if (l <= s && e <= r) return st[n];
    int mid = (s + e) / 2;
    pair<int, int> q1 = query(n + n, s, mid, l, r);
    pair<int, int> q2 = query(n + n + 1, mid + 1, e, l, r);
    return pair<int, int>(max(q1.first, q2.first), min(q1.second, q2.second));
  }
} st[32];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) scanf("%d", &a[j]);
    for (int l = 0; l < (1 << k); l++) {
      int sum = 0;
      for (int j = 0; j < k; j++) {
        if ((l >> j) & 1)
          sum += a[j];
        else
          sum -= a[j];
      }
      st[l].update(1, 1, n, i, sum);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int t, i, l, r;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d", &i);
      for (int j = 0; j < k; j++) {
        scanf("%d", &a[j]);
      }
      for (int l = 0; l < (1 << k); l++) {
        int sum = 0;
        for (int j = 0; j < k; j++) {
          if ((l >> j) & 1)
            sum += a[j];
          else
            sum -= a[j];
        }
        st[l].update(1, 1, n, i, sum);
      }
    } else {
      scanf("%d%d", &l, &r);
      int ans = -2e9;
      for (int j = 0; j < (1 << k); j++) {
        pair<int, int> x = st[j].query(1, 1, n, l, r);
        ans = max(ans, x.first - x.second);
      }
      printf("%d\n", ans);
    }
  }
}
