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 = 200000 + 1067, kk = 5, oo = 5000000;
int n, k, trai[N * 4], phai[N * 4], sum[N][1 << kk], mx[N * 4][1 << kk],
    mn[N * 4][1 << kk];
int getbit(int i, int j) { return (i >> j) & 1; }
void build(int x, int l, int r) {
  trai[x] = l;
  phai[x] = r;
  if (l == r) {
    for (int mask = 0; mask <= (1 << k) - 1; mask++)
      mx[x][mask] = mn[x][mask] = sum[l][mask];
    return;
  }
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
  for (int mask = 0; mask <= (1 << k) - 1; mask++)
    mx[x][mask] = max(mx[x * 2][mask], mx[x * 2 + 1][mask]),
    mn[x][mask] = min(mn[x * 2][mask], mn[x * 2 + 1][mask]);
}
void update(int x, int l) {
  if (trai[x] > l || phai[x] < l) return;
  if (trai[x] == phai[x]) {
    for (int mask = 0; mask <= (1 << k) - 1; mask++)
      mx[x][mask] = mn[x][mask] = sum[l][mask];
    return;
  }
  update(x * 2, l);
  update(x * 2 + 1, l);
  for (int mask = 0; mask <= (1 << k) - 1; mask++)
    mx[x][mask] = max(mx[x * 2][mask], mx[x * 2 + 1][mask]),
    mn[x][mask] = min(mn[x * 2][mask], mn[x * 2 + 1][mask]);
}
int qmx(int x, int l, int r, int mask) {
  if (trai[x] > r || phai[x] < l) return -oo;
  if (trai[x] >= l && phai[x] <= r) {
    return mx[x][mask];
  }
  return max(qmx(x * 2, l, r, mask), qmx(x * 2 + 1, l, r, mask));
}
int qmn(int x, int l, int r, int mask) {
  if (trai[x] > r || phai[x] < l) return oo;
  if (trai[x] >= l && phai[x] <= r) {
    return mn[x][mask];
  }
  return min(qmn(x * 2, l, r, mask), qmn(x * 2 + 1, l, r, mask));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  int a[5];
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k - 1; j++) cin >> a[j];
    for (int mask = 0; mask <= (1 << k) - 1; mask++) {
      sum[i][mask] = 0;
      for (int j = 0; j <= k - 1; j++)
        if (getbit(mask, j)) {
          sum[i][mask] += a[j];
        } else
          sum[i][mask] -= a[j];
    }
  }
  build(1, 1, n);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      for (int j = 0; j <= k - 1; j++) cin >> a[j];
      for (int mask = 0; mask <= (1 << k) - 1; mask++) {
        sum[x][mask] = 0;
        for (int j = 0; j <= k - 1; j++)
          if (getbit(mask, j)) {
            sum[x][mask] += a[j];
          } else
            sum[x][mask] -= a[j];
      }
      update(1, x);
    } else {
      int l, r;
      cin >> l >> r;
      int res = 0;
      for (int mask = 0; mask <= (1 << k) - 1; mask++) {
        res = max(res, qmx(1, l, r, mask) - qmn(1, l, r, mask));
      }
      cout << res << endl;
    }
  }
  return 0;
}
