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 + 10;
const int M = 1e7 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int INF = 2e9;
const int seed = 131;
int n, k, up, a[5][N];
struct node {
  int Mx[1 << 5];
  node operator+(const node &o) const {
    node res;
    for (int i = 0; i <= up; ++i) res.Mx[i] = max(Mx[i], o.Mx[i]);
    return res;
  }
} T[N << 2], tmp;
void build(int rt, int l, int r) {
  if (l == r) {
    for (int i = 0; i <= up; ++i) {
      int sum = 0;
      for (int j = 0; j < k; ++j) {
        if ((i >> j) & 1)
          sum -= a[j][r];
        else
          sum += a[j][r];
      }
      T[rt].Mx[i] = sum;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
  T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)];
}
node query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return T[rt];
  int mid = (l + r) >> 1;
  if (y <= mid)
    return query((rt << 1), l, mid, x, y);
  else if (x > mid)
    return query((rt << 1 | 1), mid + 1, r, x, y);
  else
    return query((rt << 1), l, mid, x, mid) +
           query((rt << 1 | 1), mid + 1, r, mid + 1, y);
}
void update(int rt, int l, int r, int pos) {
  if (l == r) {
    for (int i = 0; i <= up; ++i) {
      int sum = 0;
      for (int j = 0; j < k; ++j) {
        if ((i >> j) & 1)
          sum -= a[j][r];
        else
          sum += a[j][r];
      }
      T[rt].Mx[i] = sum;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    update((rt << 1), l, mid, pos);
  else
    update((rt << 1 | 1), mid + 1, r, pos);
  T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)];
}
int main() {
  scanf("%d%d", &n, &k);
  up = (1 << k) - 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) {
      scanf("%d", &a[j][i]);
    }
  }
  build(1, 1, n);
  int q, op, pos, l, r;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &pos);
      for (int j = 0; j < k; ++j) scanf("%d", &a[j][pos]);
      update(1, 1, n, pos);
    } else {
      scanf("%d%d", &l, &r);
      tmp = query(1, 1, n, l, r);
      int ans = 0;
      for (int i = 0; i <= up; ++i) ans = max(ans, tmp.Mx[i] + tmp.Mx[i ^ up]);
      printf("%d\n", ans);
    }
  }
  return 0;
}
