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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int maxn = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
int n, k;
int c[maxn], a[maxn][5];
struct Segment {
  int mx[maxn << 2];
  void build(int l, int r, int now) {
    if (l == r) {
      mx[now] = c[l];
      return;
    }
    int m = (l + r) / 2;
    build(l, m, now << 1);
    build(m + 1, r, now << 1 | 1);
    mx[now] = max(mx[now << 1], mx[now << 1 | 1]);
  }
  void modify(int x, int d, int l, int r, int now) {
    if (l == r) {
      mx[now] = d;
      return;
    }
    int m = (l + r) / 2;
    if (x <= m)
      modify(x, d, l, m, now << 1);
    else
      modify(x, d, m + 1, r, now << 1 | 1);
    mx[now] = max(mx[now << 1], mx[now << 1 | 1]);
  }
  int qmax(int x, int y, int l, int r, int now) {
    if (x <= l && r <= y) return mx[now];
    int m = (l + r) / 2, ans = -inf;
    if (x <= m) ans = max(ans, qmax(x, y, l, m, now << 1));
    if (y > m) ans = max(ans, qmax(x, y, m + 1, r, now << 1 | 1));
    return ans;
  }
} S[32];
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) a[i][j] = read();
  }
  for (int s = 0; s < 1 << k; ++s) {
    for (int i = 1; i <= n; ++i) {
      int tmp = 0;
      for (int j = 0; j < k; ++j) {
        if (s >> j & 1)
          tmp += a[i][j];
        else
          tmp -= a[i][j];
      }
      c[i] = tmp;
    }
    S[s].build(1, n, 1);
  }
  int q = read();
  while (q--) {
    int op = read();
    if (op == 1) {
      int i = read();
      for (int j = 0; j < k; ++j) a[i][j] = read();
      for (int s = 0; s < 1 << k; ++s) {
        int tmp = 0;
        for (int j = 0; j < k; ++j) {
          if (s >> j & 1)
            tmp += a[i][j];
          else
            tmp -= a[i][j];
        }
        S[s].modify(i, tmp, 1, n, 1);
      }
    } else {
      int ans = 0;
      int x = read(), y = read();
      for (int s = 0; s < 1 << k; ++s) {
        int ss = (1 << k) - 1 - s;
        ans = max(ans, S[s].qmax(x, y, 1, n, 1) + S[ss].qmax(x, y, 1, n, 1));
      }
      printf("%d\n", ans);
    }
  }
}
