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 n, k;
int a[200055][5];
struct node {
  int high[1 << 5];
  node() {
    for (int mask = 0; mask < (1 << k); mask++) {
      high[mask] = -0x3f3f3f3f;
    }
  }
  void init(int a[5]) {
    for (int mask = 0; mask < (1 << k); mask++) {
      int x = 0;
      for (int i = 0; i < k; i++) {
        if (mask & (1 << i)) {
          x += a[i];
        } else {
          x -= a[i];
        }
      }
      high[mask] = x;
    }
  }
  node merge(const node &rhs) {
    node res;
    for (int mask = 0; mask < (1 << k); mask++) {
      res.high[mask] = max(high[mask], rhs.high[mask]);
    }
    return res;
  }
  int flip(int mask) {
    for (int i = 0; i < k; i++) {
      mask ^= (1 << i);
    }
    return mask;
  }
  int query() {
    int res = 0;
    for (int mask = 0; mask < (1 << k); mask++) {
      res = max(res, high[mask] + high[flip(mask)]);
    }
    return res;
  }
};
node dat[400055];
void update(int id, int a[5]) {
  dat[id += n].init(a);
  for (; id >>= 1;) {
    dat[id] = dat[id + id].merge(dat[id + id + 1]);
  }
}
int query(int l, int r) {
  l += n, r += n;
  node res;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = res.merge(dat[l++]);
    if (r & 1) res = res.merge(dat[--r]);
  }
  return res.query();
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &a[i][j]);
    }
    dat[n + i].init(a[i]);
  }
  for (int i = n - 1; i > 0; i--) {
    dat[i] = dat[i + i].merge(dat[i + i + 1]);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int ope;
    scanf("%d", &ope);
    if (ope == 1) {
      int i;
      scanf("%d", &i);
      i--;
      int b[5];
      for (int j = 0; j < k; j++) {
        scanf("%d", &b[j]);
      }
      update(i, b);
    } else {
      int x, y;
      scanf("%d %d", &x, &y);
      x--;
      printf("%d\n", query(x, y));
    }
  }
  return 0;
}
