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 n, k, q, a[200010][10];
struct SEG {
  struct node {
    int Max, Min;
    node() {}
    node(int Max, int Min) : Max(Max), Min(Min) {}
    void init() { Max = -0x3f3f3f3f, Min = 0x3f3f3f3f; }
    node operator+(const node &other) const {
      return node(max(Max, other.Max), min(Min, other.Min));
    }
  } a[200010 << 2], res;
  void build(int id, int l, int r) {
    a[id].init();
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
  }
  void update(int id, int l, int r, int pos, int val) {
    if (l == r) {
      a[id] = node(val, val);
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      update(id << 1, l, mid, pos, val);
    else
      update(id << 1 | 1, mid + 1, r, pos, val);
    a[id] = a[id << 1] + a[id << 1 | 1];
  }
  void query(int id, int l, int r, int ql, int qr) {
    if (l >= ql && r <= qr) {
      res = res + a[id];
      return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) query(id << 1, l, mid, ql, qr);
    if (qr > mid) query(id << 1 | 1, mid + 1, r, ql, qr);
  }
} seg[1 << 5];
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j < k; ++j) scanf("%d", a[i] + j);
    for (int i = 0; i < (1 << k); ++i) {
      seg[i].build(1, 1, n);
      for (int j = 1; j <= n; ++j) {
        int tmp = 0;
        for (int o = 0; o < k; ++o)
          tmp += a[j][o] * ((((i >> o) & 1) == 0) ? -1 : 1);
        seg[i].update(1, 1, n, j, tmp);
      }
    }
    scanf("%d", &q);
    for (int i = 1, op, x, l, r; i <= q; ++i) {
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d", &x);
        for (int j = 0; j < k; ++j) scanf("%d", a[x] + j);
        for (int j = 0; j < (1 << k); ++j) {
          int tmp = 0;
          for (int o = 0; o < k; ++o)
            tmp += a[x][o] * (((j >> o) & 1) == 0 ? -1 : 1);
          seg[j].update(1, 1, n, x, tmp);
        }
      } else {
        scanf("%d%d", &l, &r);
        int res = -0x3f3f3f3f;
        for (int j = 0; j < (1 << k); ++j) {
          seg[j].res.init();
          seg[j].query(1, 1, n, l, r);
          res = max(res, seg[j].res.Max - seg[j].res.Min);
        }
        printf("%d\n", res);
      }
    }
  }
  return 0;
}
