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 maxn = 2e5 + 10;
int n, k;
pair<int, int> t[maxn << 2][1 << 5];
void update(int k, int l, int r, int pos, int x, int value) {
  if (l > pos || r < pos) return;
  if (l == r) {
    t[k][x] = make_pair(value, value);
    return;
  }
  int m = (l + r) >> 1;
  update(k << 1, l, m, pos, x, value);
  update(k << 1 | 1, m + 1, r, pos, x, value);
  t[k][x] = make_pair(max(t[k << 1][x].first, t[k << 1 | 1][x].first),
                      min(t[k << 1][x].second, t[k << 1 | 1][x].second));
}
pair<int, int> get(int k, int l, int r, int x, int i, int j) {
  if (l > j || r < i) return make_pair(-1e9, 1e9);
  if (i <= l && r <= j) return t[k][x];
  int m = (l + r) >> 1;
  pair<int, int> lo = get(k << 1, l, m, x, i, j);
  pair<int, int> hi = get(k << 1 | 1, m + 1, r, x, i, j);
  return make_pair(max(lo.first, hi.first), min(lo.second, hi.second));
}
void updateData(int id, vector<int> a) {
  for (int x = 0; x < 1 << k; ++x) {
    int value = 0;
    for (int i = 0; i < k; ++i) value += (x >> i & 1) ? -a[i] : a[i];
    update(1, 1, n, id, x, value);
  }
}
void load() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    vector<int> a(k, 0);
    for (int j = 0; j < k; ++j) scanf("%d", &a[j]);
    updateData(i, a);
  }
}
void process() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int id;
      scanf("%d", &id);
      vector<int> a(k, 0);
      for (int i = 0; i < k; ++i) scanf("%d", &a[i]);
      updateData(id, a);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int answer = 0;
      for (int x = 0; x < 1 << k; ++x) {
        pair<int, int> value = get(1, 1, n, x, l, r);
        answer = max(answer, value.first - value.second);
      }
      printf("%d\n", answer);
    }
  }
}
int main() {
  load();
  process();
  return 0;
}
