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 in[200005][5];
int seg[200005 * 4][32];
void update(int pos, int l, int r, int id, int k) {
  if (l == r) {
    for (int bit = 0; bit < (1 << k); ++bit) {
      seg[pos][bit] = 0;
      for (int j = 0; j < k; ++j) {
        if (bit & (1 << j)) {
          seg[pos][bit] += in[l][j];
        } else {
          seg[pos][bit] -= in[l][j];
        }
      }
    }
    return;
  }
  int mid = (l + r) / 2;
  if (id <= mid) {
    update(pos * 2, l, mid, id, k);
  } else {
    update(pos * 2 + 1, mid + 1, r, id, k);
  }
  for (int i = 0; i < (1 << k); ++i) {
    seg[pos][i] = max(seg[pos * 2][i], seg[pos * 2 + 1][i]);
  }
}
void query(int *maxValue, int pos, int l, int r, int ql, int qr, int k) {
  if (r < ql || qr < l) {
    return;
  }
  if (ql <= l && r <= qr) {
    for (int i = 0; i < (1 << k); ++i) {
      maxValue[i] = max(maxValue[i], seg[pos][i]);
    }
    return;
  }
  int mid = (l + r) / 2;
  query(maxValue, pos * 2, l, mid, ql, qr, k);
  query(maxValue, pos * 2 + 1, mid + 1, r, ql, qr, k);
}
void solve() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) {
      scanf("%d ", &in[i][j]);
    }
    update(1, 1, n, i, k);
  }
  int q;
  scanf("%d ", &q);
  while (q--) {
    int type;
    scanf("%d ", &type);
    if (type == 1) {
      int id;
      scanf("%d ", &id);
      for (int j = 0; j < k; ++j) {
        scanf("%d ", &in[id][j]);
      }
      update(1, 1, n, id, k);
    } else {
      int maxValue[32];
      for (int i = 0; i < 32; ++i) {
        maxValue[i] = -1e9;
      }
      int l, r;
      scanf("%d %d ", &l, &r);
      query(maxValue, 1, 1, n, l, r, k);
      int lim = (1 << k), b = lim - 1;
      int ret = 0;
      for (int i = 0; i < lim; ++i) {
        int t = b ^ i;
        ret = max(ret, maxValue[i] + maxValue[t]);
      }
      printf("%d\n", ret);
    }
  }
}
int main() {
  solve();
  return 0;
}
