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 long long mod = 998244353;
const int N = 800005;
int tr[33][N];
void update(int* tr, int rt, int l, int r, int ql, int qr, int val) {
  if (l >= ql && r <= qr) {
    tr[rt] = val;
    return;
  }
  int mid = l + r >> 1;
  if (ql <= mid) update(tr, (rt << 1), l, mid, ql, qr, val);
  if (qr > mid) update(tr, ((rt << 1) | 1), mid + 1, r, ql, qr, val);
  tr[rt] = max(tr[(rt << 1)], tr[((rt << 1) | 1)]);
}
int query(int* tr, int rt, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) {
    return tr[rt];
  }
  int mid = l + r >> 1, ans = -1e9;
  if (ql <= mid) ans = query(tr, (rt << 1), l, mid, ql, qr);
  if (qr > mid) ans = max(ans, query(tr, ((rt << 1) | 1), mid + 1, r, ql, qr));
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int en = (1 << k) - 1;
  for (int i = 0; i <= en; i++) {
    for (int j = 0; j < N; j++) {
      tr[i][j] = -1e9;
    }
  }
  for (int i = 1; i <= n; i++) {
    int a[6];
    for (int j = 1; j <= k; j++) {
      scanf("%d", &a[j]);
    }
    int enn = en / 2;
    for (int j = 0; j <= enn; j++) {
      int val = 0;
      int oppo = 0;
      for (int bit = 1, ct = 1; ct <= k; bit <<= 1, ct++) {
        if (j & bit)
          val += a[ct];
        else
          val -= a[ct], oppo |= bit;
      }
      update(tr[j], 1, 1, n, i, i, val);
      update(tr[oppo], 1, 1, n, i, i, -val);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int id, a[6];
      scanf("%d", &id);
      for (int i = 1; i <= k; i++) scanf("%d", a + i);
      int enn = en;
      for (int j = 0; j <= enn; j++) {
        int oppo = 0;
        int val = 0;
        for (int bit = 1, ct = 1; ct <= k; bit <<= 1, ct++) {
          if (j & bit)
            val += a[ct];
          else
            val -= a[ct], oppo |= bit;
        }
        update(tr[j], 1, 1, n, id, id, val);
        update(tr[oppo], 1, 1, n, id, id, -val);
      }
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = -1e9;
      int enn = en / 2;
      for (int j = 0; j <= enn; j++) {
        int oppo = 0;
        for (int bit = 1, ct = 1; ct <= k; bit <<= 1, ct++) {
          if (!(j & bit)) oppo |= bit;
        }
        ans = max(ans,
                  query(tr[j], 1, 1, n, l, r) + query(tr[oppo], 1, 1, n, l, r));
      }
      printf("%d\n", ans);
    }
  }
}
