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, t[2][32][400005], p[400005][5], q;
void init(int i) {
  for (int j = 0; j < 1 << k; j++) {
    int te = j;
    t[0][j][i + n] = 0;
    t[1][j][i + n] = 0;
    for (int l = 0; l < k; l++) {
      if (te % 2 == 0) {
        t[0][j][i + n] -= p[i][l];
        t[1][j][i + n] -= p[i][l];
      } else {
        t[0][j][i + n] += p[i][l];
        t[1][j][i + n] += p[i][l];
      }
      te /= 2;
    }
  }
}
void build() {
  for (int i = 0; i < n; i++) {
    init(i);
  }
  for (int i = n - 1; i > 0; i--) {
    for (int j = 0; j < 1 << k; j++) {
      t[0][j][i] = max(t[0][j][2 * i], t[0][j][2 * i + 1]);
      t[1][j][i] = min(t[1][j][2 * i], t[1][j][2 * i + 1]);
    }
  }
  return;
}
void update(int x) {
  init(x);
  for (x += n; x > 1; x /= 2) {
    for (int j = 0; j < 1 << k; j++) {
      t[0][j][x / 2] = max(t[0][j][x], t[0][j][x ^ 1]);
      t[1][j][x / 2] = min(t[1][j][x], t[1][j][x ^ 1]);
    }
  }
  return;
}
int query(int l, int r) {
  int ans = 0;
  l += n;
  r += n;
  for (int i = 0; i < 1 << k; i++) {
    int l0 = l, r0 = r, mx = -2e9, mn = 2e9;
    for (; l0 < r0; l0 /= 2, r0 /= 2) {
      if (l0 & 1) {
        mx = max(mx, t[0][i][l0]);
        mn = min(mn, t[1][i][l0]);
        l0++;
      }
      if (r0 & 1) {
        mx = max(mx, t[0][i][r0 - 1]);
        mn = min(mn, t[1][i][r0 - 1]);
        r0--;
      }
    }
    ans = max(ans, mx - mn);
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &p[i][j]);
    }
  }
  build();
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a;
    scanf("%d", &a);
    if (a == 1) {
      int ind;
      scanf("%d", &ind);
      ind--;
      for (int j = 0; j < k; j++) {
        scanf("%d", &p[ind][j]);
      }
      update(ind);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      cout << query(l - 1, r) << endl;
    }
  }
}
