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;
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
using ll = long long;
using ld = long double;
const int N = 2e5 + 7;
const ll mod = 1e9 + 7;
int n, m;
vector<int> a[N];
template <class T>
struct msegtree {
  T tree[4 * N];
  T eval(T x, T y) { return min(x, y); }
  void update(int ix, T val, int node = 1, int L = 0, int R = N - 1) {
    if (L == R) return void(tree[node] = val);
    int mid = (L + R) >> 1;
    if (ix <= mid)
      update(ix, val, node * 2, L, mid);
    else
      update(ix, val, node * 2 + 1, mid + 1, R);
    tree[node] = eval(tree[node * 2], tree[node * 2 + 1]);
  }
  T query(int l, int r, int node = 1, int L = 0, int R = N - 1) {
    if (l > r || l > R || r < L) return 1e9;
    if (L >= l && R <= r) return tree[node];
    int mid = (L + R) >> 1;
    return eval(query(l, r, node * 2, L, mid),
                query(l, r, node * 2 + 1, mid + 1, R));
  }
};
template <class T>
struct Msegtree {
  T tree[4 * N];
  T eval(T x, T y) { return max(x, y); }
  void update(int ix, T val, int node = 1, int L = 0, int R = N - 1) {
    if (L == R) return void(tree[node] = val);
    int mid = (L + R) >> 1;
    if (ix <= mid)
      update(ix, val, node * 2, L, mid);
    else
      update(ix, val, node * 2 + 1, mid + 1, R);
    tree[node] = eval(tree[node * 2], tree[node * 2 + 1]);
  }
  T query(int l, int r, int node = 1, int L = 0, int R = N - 1) {
    if (l > r || l > R || r < L) return -1e9;
    if (L >= l && R <= r) return tree[node];
    int mid = (L + R) >> 1;
    return eval(query(l, r, node * 2, L, mid),
                query(l, r, node * 2 + 1, mid + 1, R));
  }
};
msegtree<int> Mins[32];
Msegtree<int> Maxs[32];
void update(int ix) {
  for (int j = 0; j < (1 << m); ++j) {
    int sum = 0;
    for (int k = 0; k < m; ++k) sum += ((j & (1 << k)) ? a[ix][k] : -a[ix][k]);
    Maxs[j].update(ix, sum);
    Mins[j].update(ix, sum);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    vector<int> v(m);
    for (auto &u : v) cin >> u;
    a[i] = v;
    update(i);
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int ix;
      cin >> ix;
      --ix;
      vector<int> v(m);
      for (auto &u : v) cin >> u;
      a[ix] = v;
      update(ix);
    } else {
      int l, r;
      cin >> l >> r;
      --l, --r;
      int ans = -1e9;
      for (int j = 0; j < 32; ++j) {
        int ret = Maxs[j].query(l, r) - Mins[j].query(l, r);
        ans = max(ans, ret);
      }
      cout << ans << "\n";
    }
  }
  return 0;
}
