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 a[200005][40];
int n, k;
int Minn[200005 * 4][40], Maxn[200005 * 4][40];
int Minnn, Maxnn, ans;
void modify(int t, int l, int r, int pos, int s) {
  if (l == r) {
    int sum = 0;
    for (int i = 0; i < k; i++) {
      if ((1 << i) & s)
        sum += a[l][i];
      else
        sum -= a[l][i];
    }
    Minn[t][s] = Maxn[t][s] = sum;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid) {
    modify(t * 2, l, mid, pos, s);
  } else {
    modify(t * 2 + 1, mid + 1, r, pos, s);
  }
  Maxn[t][s] = max(Maxn[t * 2][s], Maxn[t * 2 + 1][s]);
}
void dfs(int t, int s, int pos) {
  if (t == k) {
    modify(1, 1, n, pos, s);
  } else {
    dfs(t + 1, s, pos);
    dfs(t + 1, s + (1 << t), pos);
  }
}
void query(int t, int l, int r, int L, int R, int s1, int s2) {
  if (L <= l && r <= R) {
    Minnn = max(Minnn, Maxn[t][s2]);
    Maxnn = max(Maxnn, Maxn[t][s1]);
    return;
  }
  int mid = (l + r) / 2;
  if (L <= mid) query(t * 2, l, mid, L, R, s1, s2);
  if (mid < R) query(t * 2 + 1, mid + 1, r, L, R, s1, s2);
}
void build(int t, int l, int r) {
  for (int j = 0; j < (1 << k); j++) {
    Maxn[t][j] = -10000000;
  }
  if (l <= r) {
    return;
  }
  int mid = (l + r) / 2;
  build(t * 2, l, mid);
  build(t * 2 + 1, mid + 1, r);
}
int main() {
  scanf("%d%d", &n, &k);
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) scanf("%d", &a[i][j]);
    dfs(0, 0, i);
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    int Ca;
    scanf("%d", &Ca);
    if (Ca == 1) {
      int pos;
      scanf("%d", &pos);
      for (int i = 0; i < k; i++) scanf("%d", &a[pos][i]);
      dfs(0, 0, pos);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      ans = -10000000;
      for (int i = 0; i < (1 << k); i++) {
        Minnn = -10000000;
        Maxnn = -10000000;
        query(1, 1, n, l, r, i, (((1 << k) - 1) ^ i));
        ans = max(ans, Minnn + Maxnn);
      }
      printf("%d\n", ans);
    }
  }
}
