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 tr[200005 << 2][32];
int a[200005][5];
int n, k;
void pushup(int rt) {
  for (int i = 0; i < (1 << k); i++) {
    tr[rt][i] = min(tr[rt << 1][i], tr[rt << 1 | 1][i]);
  }
}
void build(int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      int tmp = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1) {
          tmp += a[l][j];
        } else {
          tmp -= a[l][j];
        }
      }
      tr[rt][i] = tmp;
    }
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void update(int l, int r, int rt, int x) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      int tmp = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1) {
          tmp += a[l][j];
        } else {
          tmp -= a[l][j];
        }
      }
      tr[rt][i] = tmp;
    }
    return;
  }
  int m = (l + r) >> 1;
  if (x <= m)
    update(l, m, rt << 1, x);
  else
    update(m + 1, r, rt << 1 | 1, x);
  pushup(rt);
}
int query(int l, int r, int rt, int L, int R, int f) {
  if (L <= l && R >= r) {
    return tr[rt][f];
  }
  int m = (l + r) >> 1;
  int ans = 1e9;
  if (L <= m) ans = min(ans, query(l, m, rt << 1, L, R, f));
  if (R > m) ans = min(ans, query(m + 1, r, rt << 1 | 1, L, R, f));
  return ans;
}
void work() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  build(1, n, 1);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x;
      scanf("%d", &x);
      for (int i = 0; i < k; i++) {
        scanf("%d", &a[x][i]);
      }
      update(1, n, 1, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int i = 0; i < (1 << k); i++) {
        int x = query(1, n, 1, l, r, i);
        int y = query(1, n, 1, l, r, ((1 << k) - 1) ^ i);
        ans = max(ans, abs(x + y));
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    work();
  }
}
