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;
struct seg_tree {
  int num[200005];
  int dmin[800005];
  int dmax[800005];
  void build(int i, int l, int r) {
    if (l < r) {
      build(i << 1, l, l + r >> 1);
      build(i << 1 | 1, (l + r >> 1) + 1, r);
      push_up(i);
    } else {
      dmax[i] = dmin[i] = num[l];
    }
  }
  inline void push_up(int i) {
    dmax[i] = max(dmax[i << 1], dmax[i << 1 | 1]);
    dmin[i] = min(dmin[i << 1], dmin[i << 1 | 1]);
  }
  pair<int, int> query(int i, int l, int r, int l1, int r1) {
    if (l == l1 && r == r1)
      return {dmax[i], dmin[i]};
    else {
      int mid = l + r >> 1;
      if (l1 > mid) {
        return query(i << 1 | 1, mid + 1, r, l1, r1);
      } else if (r1 <= mid) {
        return query(i << 1, l, mid, l1, r1);
      } else {
        pair<int, int> tmp, tmp2;
        tmp = query(i << 1, l, mid, l1, mid);
        tmp2 = query(i << 1 | 1, mid + 1, r, mid + 1, r1);
        return {max(tmp.first, tmp2.first), min(tmp.second, tmp2.second)};
      }
    }
  }
  inline void updat(int i, int l, int r, int l1, int x) {
    if (l == l1 && r == l1)
      dmax[i] = dmin[i] = x;
    else {
      int mid = l + r >> 1;
      if (l1 > mid)
        updat(i << 1 | 1, mid + 1, r, l1, x);
      else if (l1 <= mid)
        updat(i << 1, l, mid, l1, x);
      push_up(i);
    }
  }
} tr[1 << 5];
int p[200050][5];
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", &p[i][j]);
    }
    for (int j = 0; j < (1 << k); j++) {
      int sum = 0;
      for (int z = 0; z < k; z++) {
        if ((1 << z) & j) {
          sum += p[i][z];
        } else {
          sum -= p[i][z];
        }
      }
      tr[j].num[i] = sum;
    }
  }
  for (int j = 0; j < (1 << k); j++) {
    tr[j].build(1, 1, n);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int index;
      scanf("%d", &index);
      for (int i = 0; i < k; i++) {
        scanf("%d", &p[index][i]);
      }
      for (int j = 0; j < (1 << k); j++) {
        int sum = 0;
        for (int z = 0; z < k; z++) {
          if ((1 << z) & j) {
            sum += p[index][z];
          } else {
            sum -= p[index][z];
          }
        }
        tr[j].updat(1, 1, n, index, sum);
      }
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int j = 0; j < (1 << k); j++) {
        pair<int, int> tmp = tr[j].query(1, 1, n, l, r);
        ans = max(ans, tmp.first - tmp.second);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
