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 = 2e5 + 50;
struct Node {
  vector<int> val;
  Node() { val.resize(35); }
} node[N << 2];
int n, k, q;
int a[N][6];
void Max(const vector<int> &x, const vector<int> &y, vector<int> &z) {
  for (int i = 0; i < (1 << k); ++i) z[i] = max(x[i], y[i]);
}
void Max(const vector<int> &x, vector<int> &y) {
  for (int i = 0; i < (1 << k); ++i) y[i] = max(x[i], y[i]);
}
void build(int u, int l, int r) {
  if (l == r) {
    for (int s = 0; s < (1 << k); ++s) {
      for (int t = 1; t <= k; ++t) {
        if (s & (1 << (t - 1)))
          node[u].val[s] += a[l][t];
        else
          node[u].val[s] -= a[l][t];
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (l <= mid) build((u << 1), l, mid);
  if (r > mid) build((u << 1 | 1), mid + 1, r);
  Max(node[(u << 1)].val, node[(u << 1 | 1)].val, node[u].val);
}
void modify(int u, int l, int r, int pos) {
  if (l == r) {
    for (int i = 0; i < (1 << k); ++i) node[u].val[i] = 0;
    for (int s = 0; s < (1 << k); ++s) {
      for (int t = 1; t <= k; ++t) {
        if (s & (1 << (t - 1)))
          node[u].val[s] += a[l][t];
        else
          node[u].val[s] -= a[l][t];
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify((u << 1), l, mid, pos);
  else
    modify((u << 1 | 1), mid + 1, r, pos);
  Max(node[(u << 1)].val, node[(u << 1 | 1)].val, node[u].val);
}
vector<int> ans(35);
void query(int u, int l, int r, int ql, int qr) {
  if (ql <= l && qr >= r) {
    Max(node[u].val, ans);
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) query((u << 1), l, mid, ql, qr);
  if (qr > mid) query((u << 1 | 1), mid + 1, r, ql, qr);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i)
    for (int t = 1; t <= k; ++t) scanf("%d", &a[i][t]);
  build(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int opt, id, l, r;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d", &id);
      for (int i = 1; i <= k; ++i) scanf("%d", &a[id][i]);
      modify(1, 1, n, id);
    } else {
      for (int i = 0; i < 35; ++i) ans[i] = -1e9;
      scanf("%d%d", &l, &r);
      query(1, 1, n, l, r);
      int maxx = 0;
      for (int s = 0; s < (1 << k); ++s) {
        maxx = max(maxx, ans[s] + ans[(1 << k) - 1 - s]);
      }
      printf("%d\n", maxx);
    }
  }
}
