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 mod = 998244353, MAX = 1 << 18, INF = 1000000000;
int n;
long long dat[2 * MAX - 1][64];
void init(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < 2 * n - 1; i++) {
    for (int j = 0; j < 64; j++) {
      if (j < 32)
        dat[i][j] = -INF;
      else
        dat[i][j] = INF;
    }
  }
}
void update(int k, vector<long long> &a) {
  k += n - 1;
  for (int j = 0; j < 32; j++) {
    int sum = 0;
    for (int x = 0; x < 5; x++) {
      if (j & (1 << x))
        sum += a[x];
      else
        sum -= a[x];
    }
    dat[k][j] = sum;
    dat[k][j + 32] = sum;
  }
  while (k > 0) {
    k = (k - 1) / 2;
    for (int j = 0; j < 64; j++) {
      if (j < 32) {
        dat[k][j] = max(dat[k * 2 + 1][j], dat[k * 2 + 2][j]);
      } else {
        dat[k][j] = min(dat[k * 2 + 1][j], dat[k * 2 + 2][j]);
      }
    }
  }
}
long long query(int a, int b, int k, int l, int r, int kind) {
  if (kind < 32) {
    if (r <= a || b <= l) return -INF;
    if (a <= l && r <= b)
      return dat[k][kind];
    else {
      long long vl = query(a, b, 2 * k + 1, l, (l + r) / 2, kind);
      long long vr = query(a, b, 2 * k + 2, (l + r) / 2, r, kind);
      return max(vl, vr);
    }
  } else {
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b)
      return dat[k][kind];
    else {
      long long vl = query(a, b, 2 * k + 1, l, (l + r) / 2, kind);
      long long vr = query(a, b, 2 * k + 2, (l + r) / 2, r, kind);
      return min(vl, vr);
    }
  }
}
int main() {
  int N, K;
  cin >> N >> K;
  init(N);
  for (int i = 0; i < N; i++) {
    vector<long long> A(5, 0);
    for (int k = 0; k < K; k++) {
      scanf("%lld", &A[k]);
    }
    update(i, A);
  }
  int Q;
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    int x;
    scanf("%d", &x);
    if (x == 1) {
      int y;
      scanf("%d", &y);
      y--;
      vector<long long> A(5, 0);
      for (int k = 0; k < K; k++) {
        scanf("%lld", &A[k]);
      }
      update(y, A);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      r--;
      long long ans = -2 * INF;
      for (int j = 0; j < 32; j++) {
        long long a = query(l, r + 1, 0, 0, n, j);
        long long b = query(l, r + 1, 0, 0, n, 32 + j);
        ans = max(ans, a - b);
      }
      cout << ans << "\n";
    }
  }
}
