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>
const int N = 200200;
const long long mod = 998244353;
using namespace std;
int n;
int k;
vector<int> a[N];
int t1[4 * N][1 << 5];
int t2[4 * N][1 << 5];
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      int s = 0;
      for (int j = 0; j < k; j++) {
        if (i & (1 << j)) {
          s += a[l][j];
        } else {
          s -= a[l][j];
        }
      }
      t1[x][i] = t2[x][i] = s;
    }
    return;
  }
  int m = (l + r) / 2;
  build(x * 2, l, m);
  build(x * 2 + 1, m + 1, r);
  for (int i = 0; i < (1 << k); i++) {
    t1[x][i] = min(t1[x * 2][i], t1[x * 2 + 1][i]);
    t2[x][i] = max(t2[x * 2][i], t2[x * 2 + 1][i]);
  }
}
void upd(int x, int l, int r, int g) {
  if (l == r) {
    for (int i = 0; i < (1 << k); i++) {
      int s = 0;
      for (int j = 0; j < k; j++) {
        if (i & (1 << j)) {
          s += a[l][j];
        } else {
          s -= a[l][j];
        }
      }
      t1[x][i] = t2[x][i] = s;
    }
    return;
  }
  int m = (l + r) / 2;
  if (g <= m) {
    upd(x * 2, l, m, g);
  } else {
    upd(x * 2 + 1, m + 1, r, g);
  }
  for (int i = 0; i < (1 << k); i++) {
    t1[x][i] = min(t1[x * 2][i], t1[x * 2 + 1][i]);
    t2[x][i] = max(t2[x * 2][i], t2[x * 2 + 1][i]);
  }
}
int a1[1 << 5];
int a2[1 << 5];
void get(int x, int l, int r, int tl, int tr) {
  if (tl > tr) {
    return;
  }
  if (l == tl && r == tr) {
    for (int i = 0; i < (1 << 5); i++) {
      a1[i] = min(a1[i], t1[x][i]);
      a2[i] = max(a2[i], t2[x][i]);
    }
    return;
  }
  int m = (l + r) / 2;
  get(x * 2, l, m, tl, min(m, tr));
  get(x * 2 + 1, m + 1, r, max(m + 1, tl), tr);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      int x;
      scanf("%d", &x);
      a[i].push_back(x);
    }
  }
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  int ms = (1 << k) - 1;
  for (int i = 1; i <= q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      int l, r;
      scanf("%d%d", &l, &r);
      for (int j = 0; j < (1 << k); j++) {
        a1[j] = 1e9;
        a2[j] = -1e9;
      }
      get(1, 1, n, l, r);
      int best = 0;
      for (int j = 0; j < (1 << k); j++) {
        best = max(best, abs(a1[j] - a2[j]));
      }
      printf("%d\n", best);
    } else {
      int x;
      scanf("%d", &x);
      for (int j = 0; j < k; j++) {
        int y;
        scanf("%d", &y);
        a[x][j] = y;
      }
      upd(1, 1, n, x);
    }
  }
}
