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 all, n, k, m = 1, qq, ty, p, l, r;
long long int mi[32][524288], ma[32][524288], a[524288][5];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  while (m < n) m <<= 1;
  all = 1 << k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < k; j++) cin >> a[i][j];
  for (int b = 0; b < all; b++) {
    for (int i = 0; i < n; i++) {
      long long int val = 0;
      for (int j = 0; j < k; j++) {
        if (((b >> j) & 1) == 1)
          val += a[i][j];
        else
          val -= a[i][j];
      }
      mi[b][i + m] = val;
      ma[b][i + m] = val;
    }
    for (int i = n; i < m; i++) {
      mi[b][i + m] = 1000000000;
      ma[b][i + m] = -1000000000;
    }
    for (int i = m - 1; i > 0; i--) {
      mi[b][i] = min(mi[b][i << 1], mi[b][i << 1 | 1]);
      ma[b][i] = max(ma[b][i << 1], ma[b][i << 1 | 1]);
    }
  }
  cin >> qq;
  while (qq-- > 0) {
    cin >> ty;
    if (ty == 1) {
      cin >> p;
      p--;
      for (int i = 0; i < k; i++) cin >> a[p][i];
      for (int b = 0; b < all; b++) {
        long long int val = 0;
        for (int j = 0; j < k; j++) {
          if (((b >> j) & 1) == 1)
            val += a[p][j];
          else
            val -= a[p][j];
        }
        mi[b][p + m] = val;
        ma[b][p + m] = val;
        for (int pp = (p + m) >> 1; pp > 0; pp >>= 1) {
          mi[b][pp] = min(mi[b][pp << 1], mi[b][pp << 1 | 1]);
          ma[b][pp] = max(ma[b][pp << 1], ma[b][pp << 1 | 1]);
        }
      }
    } else {
      cin >> l >> r;
      l--;
      r--;
      long long int ans = 0;
      for (int b = 0; b < all; b++) {
        long long int mx = -1000000000, mn = 1000000000;
        for (int ll = l + m, rr = r + m + 1; ll < rr; ll >>= 1, rr >>= 1) {
          if (ll & 1) {
            mx = max(mx, ma[b][ll]);
            mn = min(mn, mi[b][ll]);
            ll++;
          }
          if (rr & 1) {
            rr--;
            mx = max(mx, ma[b][rr]);
            mn = min(mn, mi[b][rr]);
          }
        }
        ans = max(ans, mx - mn);
      }
      cout << ans << "\n";
    }
  }
}
