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 int maxn = 2e5 + 5;
const int N = 1e9 + 1;
int k, n;
int a[maxn][5];
int tmp[5];
struct SEG {
  int state;
  int mi[maxn << 2], ma[maxn << 2];
  void push_up(int rt) {
    mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]);
    ma[rt] = max(ma[rt << 1], ma[rt << 1 | 1]);
  }
  void build(int rt, int l, int r) {
    if (l == r) {
      int val = 0;
      for (int i = 0; i < k; i++) {
        if ((state & (1 << i)) > 0)
          val += a[l][i];
        else
          val -= a[l][i];
      }
      mi[rt] = ma[rt] = val;
      return;
    }
    int m = (l + r) >> 1;
    build(rt << 1, l, m);
    build(rt << 1 | 1, m + 1, r);
    push_up(rt);
  }
  void update(int rt, int pos, int l, int r) {
    if (l == r) {
      int val = 0;
      for (int i = 0; i < k; i++) {
        if ((state & (1 << i)) > 0)
          val += tmp[i];
        else
          val -= tmp[i];
      }
      mi[rt] = ma[rt] = val;
      return;
    }
    int m = (l + r) >> 1;
    if (m >= pos)
      update(rt << 1, pos, l, m);
    else
      update(rt << 1 | 1, pos, m + 1, r);
    push_up(rt);
  }
  pair<int, int> query(int rt, int L, int R, int l, int r) {
    if (L <= l && r <= R) return {mi[rt], ma[rt]};
    int m = (l + r) >> 1;
    pair<int, int> ret1 = {1e8, -1e8}, ret2 = {1e8, -1e8};
    if (m >= L) ret1 = query(rt << 1, L, R, l, m);
    if (m < R) ret2 = query(rt << 1 | 1, L, R, m + 1, r);
    return {min(ret1.first, ret2.first), max(ret1.second, ret2.second)};
  }
  int ask(int L, int R) {
    pair<int, int> x = query(1, L, R, 1, n);
    return x.second - x.first;
  }
} tr[32];
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) scanf("%d", &a[i][j]);
  }
  for (int i = 0; i < (1 << k); i++) {
    tr[i].state = i;
    tr[i].build(1, 1, n);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos;
      scanf("%d", &pos);
      for (int i = 0; i < k; i++) scanf("%d", tmp + i);
      for (int i = 0; i < (1 << k); i++) tr[i].update(1, pos, 1, n);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int i = 0; i < (1 << k); i++) ans = max(ans, tr[i].ask(l, r));
      printf("%d\n", ans);
    }
  }
}
