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 N = (int)2e5 + 123, inf = 1e9;
const long long INF = 1e18;
int n, k, q;
struct point {
  int x[(1 << 5)];
  point() {}
  void read() {
    for (int i = 0; i < k; i++) scanf("%d", &x[i]);
  }
  int dist(point a) {
    int res = 0;
    for (int i = 0; i < k; i++) res += abs(x[i] - a.x[i]);
    return res;
  }
  int get(int mask) {
    int res = 0;
    for (int i = 0; i < k; i++)
      if ((1 << i) & mask)
        res += x[i];
      else
        res -= x[i];
    return res;
  }
} a[N];
struct btr {
  int val;
  int mx[(1 << 5)];
  btr() { val = -1; }
  btr(int id) {
    val = 0;
    for (int i = 0; i < (1 << k); i++) mx[i] = a[id].get(i);
  }
} t[N << 2];
btr Merge(btr l, btr r) {
  if (l.val == -1) return r;
  if (r.val == -1) return l;
  btr res;
  res.val = max(l.val, r.val);
  for (int i = 0; i < (1 << k); i++)
    res.val = max(res.val, l.mx[i] + r.mx[(1 << k) - 1 - i]);
  for (int i = 0; i < (1 << k); i++) res.mx[i] = max(l.mx[i], r.mx[i]);
  return res;
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = btr(tl);
    return;
  }
  int tm = (tl + tr) >> 1;
  build(v << 1, tl, tm);
  build(v << 1 | 1, tm + 1, tr);
  t[v] = Merge(t[v << 1], t[v << 1 | 1]);
}
void upd(int v, int tl, int tr, int pos) {
  if (tl == tr) {
    t[v] = btr(tl);
    return;
  }
  int tm = (tl + tr) >> 1;
  if (pos <= tm)
    upd(v << 1, tl, tm, pos);
  else
    upd(v << 1 | 1, tm + 1, tr, pos);
  t[v] = Merge(t[v << 1], t[v << 1 | 1]);
}
btr get(int v, int tl, int tr, int l, int r) {
  if (r < tl || tr < l || l > r) return btr();
  if (l <= tl && tr <= r) return t[v];
  int tm = (tl + tr) >> 1;
  return Merge(get(v << 1, tl, tm, l, r), get(v << 1 | 1, tm + 1, tr, l, r));
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) a[i].read();
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int id;
      scanf("%d", &id);
      a[id].read();
      upd(1, 1, n, id);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int res = get(1, 1, n, l, r).val;
      printf("%d\n", res);
    }
  }
  return 0;
}
