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 + 5;
const int inf = 1e9;
int Set(int N, int pos) { return N |= (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
int n, d;
int a[N][10];
int tree[4 * N][(1 << 4) + 5][2];
void build(int node, int b, int e, int id) {
  if (b > e) return;
  if (b == e) {
    for (int i = 0; i < d; i++) {
      if (Check(id, i)) {
        tree[node][id][0] += a[b][i];
        tree[node][id][1] += a[b][i];
      } else {
        tree[node][id][0] -= a[b][i];
        tree[node][id][1] -= a[b][i];
      }
    }
    return;
  }
  int l = node << 1, r = l | 1, m = (b + e) >> 1;
  build(l, b, m, id);
  build(r, m + 1, e, id);
  tree[node][id][1] = max(tree[l][id][1], tree[r][id][1]);
  tree[node][id][0] = min(tree[l][id][0], tree[r][id][0]);
}
void update(int node, int b, int e, int pos, int id) {
  if (pos > e || pos < b || b > e) return;
  if (b == e && pos == b) {
    tree[node][id][0] = 0;
    tree[node][id][1] = 0;
    for (int i = 0; i < d; i++) {
      if (Check(id, i)) {
        tree[node][id][0] += a[b][i];
        tree[node][id][1] += a[b][i];
      } else {
        tree[node][id][0] -= a[b][i];
        tree[node][id][1] -= a[b][i];
      }
    }
    return;
  }
  int l = node << 1, r = l | 1, m = (b + e) >> 1;
  update(l, b, m, pos, id);
  update(r, m + 1, e, pos, id);
  tree[node][id][1] = max(tree[l][id][1], tree[r][id][1]);
  tree[node][id][0] = min(tree[l][id][0], tree[r][id][0]);
}
pair<int, int> query(int node, int b, int e, int i, int j, int id) {
  if (i > e || j < b) return {-inf, inf};
  if (i <= b && j >= e) return {tree[node][id][1], tree[node][id][0]};
  int l = node << 1, r = l | 1, m = (b + e) >> 1;
  pair<int, int> ret = {-inf, inf};
  pair<int, int> L = query(l, b, m, i, j, id);
  pair<int, int> R = query(r, m + 1, e, i, j, id);
  ret.first = max(ret.first, max(L.first, R.first));
  ret.second = min(ret.second, min(L.second, R.second));
  return ret;
}
int main(int argc, char const *argv[]) {
  scanf("%d %d", &n, &d);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < d; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  for (int i = 0; i < (1 << (d - 1)); i++) build(1, 1, n, i);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int pos;
      scanf("%d", &pos);
      for (int i = 0; i < d; i++) scanf("%d", &a[pos][i]);
      for (int i = 0; i < (1 << (d - 1)); i++) update(1, 1, n, pos, i);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      int ans = -inf;
      for (int i = 0; i < (1 << (d - 1)); i++) {
        pair<int, int> p = query(1, 1, n, l, r, i);
        ans = max(ans, abs(p.first - p.second));
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
