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 SIZE = 1 << 18;
int a[200005][5], b[5], k;
struct ST {
  pair<long long, int> data1[2 * SIZE], data2[2 * SIZE];
  void update(int x, long long v) {
    x += SIZE;
    data1[x] = {v, x - SIZE};
    data2[x] = {v, x - SIZE};
    for (x >>= 1; x; x >>= 1) {
      data2[x] = min(data2[x << 1], data2[x << 1 | 1]);
      data1[x] = max(data1[x << 1], data1[x << 1 | 1]);
    }
  }
  int query1(int s, int e) {
    pair<long long, int> ret = {-1e18, 0};
    for (s += SIZE, e += SIZE; s <= e; s >>= 1, e >>= 1) {
      if (s & 1) ret = max(ret, data1[s++]);
      if (~e & 1) ret = max(ret, data1[e--]);
    }
    return ret.second;
  }
  int query2(int s, int e) {
    pair<long long, int> ret = {1e18, 0};
    for (s += SIZE, e += SIZE; s <= e; s >>= 1, e >>= 1) {
      if (s & 1) ret = min(ret, data2[s++]);
      if (~e & 1) ret = min(ret, data2[e--]);
    }
    return ret.second;
  }
} seg[16];
long long dist(int i, int j) {
  long long ret = 0;
  for (int x = 0; x < k; ++x) ret += abs(a[i][x] - a[j][x]);
  return ret;
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < k; ++j) scanf("%d", a[i] + j);
  }
  for (int j = 1; j <= n; ++j) {
    for (int mask = (1 << (k - 1)) - 1; mask >= 0; --mask) {
      long long v = 0;
      for (int i = 0; i < k; ++i) {
        if ((mask >> i) & 1)
          v += a[j][i];
        else
          v -= a[j][i];
      }
      seg[mask].update(j, v);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int x;
      scanf("%d", &x);
      for (int i = 0; i < k; ++i) scanf("%d", a[x] + i);
      for (int mask = (1 << (k - 1)) - 1; mask >= 0; --mask) {
        long long v = 0;
        for (int i = 0; i < k; ++i) {
          if ((mask >> i) & 1)
            v += a[x][i];
          else
            v -= a[x][i];
        }
        seg[mask].update(x, v);
      }
    } else {
      vector<int> vec;
      int l, r;
      scanf("%d%d", &l, &r);
      for (int i = (1 << (k - 1)) - 1; i >= 0; --i) {
        vec.push_back(seg[i].query1(l, r));
        vec.push_back(seg[i].query2(l, r));
      }
      long long ans = 0;
      for (int i = 0; i < vec.size(); ++i) {
        for (int j = 0; j < i; ++j) {
          ans = max(ans, dist(vec[i], vec[j]));
        }
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
