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;
struct point {
  int a[6];
  point(){};
  int &operator[](int x) { return a[x]; }
};
int n, k, q;
int tree[35][800005];
point p[200005];
int sign(point &p, int bit) {
  int ans = 0;
  for (int i = 0; i < k; i++) {
    if (bit & 1 == 1) {
      ans += p[i];
    } else {
      ans -= p[i];
    }
    bit >>= 1;
  }
  return ans;
}
void build_it(int t, int l, int r) {
  if (l == r - 1) {
    for (int i = 0; i < (1 << k); i++) {
      tree[i][t] = sign(p[l], i);
    }
    return;
  }
  int mid = (l + r) / 2;
  build_it(t * 2, l, mid);
  build_it(t * 2 + 1, mid, r);
  for (int i = 0; i < (1 << k); i++) {
    tree[i][t] = min(tree[i][t * 2], tree[i][t * 2 + 1]);
  }
}
void update_it(int t, int l, int r, int pos, point &val) {
  if (l == r - 1) {
    for (int i = 0; i < (1 << k); i++) {
      tree[i][t] = sign(val, i);
    }
    return;
  }
  int mid = (l + r) / 2;
  if (pos < mid) {
    update_it(t * 2, l, mid, pos, val);
  } else {
    update_it(t * 2 + 1, mid, r, pos, val);
  }
  for (int i = 0; i < (1 << k); i++) {
    tree[i][t] = min(tree[i][t * 2], tree[i][t * 2 + 1]);
  }
}
int good[35];
void getans(int t, int l, int r, int a, int b) {
  if (a >= b) return;
  if (l == a && r == b) {
    for (int i = 0; i < (1 << k); i++) {
      good[i] = min(good[i], tree[i][t]);
    }
    return;
  }
  int mid = (l + r) / 2;
  getans(t * 2, l, mid, a, min(mid, b));
  getans(t * 2 + 1, mid, r, max(mid, a), b);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      scanf("%d", &p[i][j]);
    }
  }
  build_it(1, 0, n);
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int x;
      scanf("%d", &x);
      x--;
      point tmp;
      for (int i = 0; i < k; i++) {
        scanf("%d", &tmp[i]);
      }
      update_it(1, 0, n, x, tmp);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      for (int i = 0; i < (1 << k); i++) {
        good[i] = 1e9 + 7;
      }
      getans(1, 0, n, l, r);
      int res = 0;
      for (int i = 0; i < (1 << k); i++) {
        res = max(res, abs(good[(1 << k) - i - 1] + good[i]));
      }
      printf("%d\n", res);
    }
  }
}
