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;
const int K = 5;
const int INF = 1e9;
int a[N + 7][K + 2];
int qt314[K + 2];
struct SEG {
  int tree[4 * N + 7][2];
  int query(int ss, int se, int si, int qs, int qe, bool fl) {
    if (qs > qe || ss > qe || qs > se) {
      return fl == 0 ? INF : -INF;
    }
    if (qs <= ss && se <= qe) {
      return tree[si][fl];
    }
    int t1 = query(ss, ((ss + se) >> 1), si * 2, qs, qe, fl);
    int t2 = query(((ss + se) >> 1) + 1, se, si * 2 + 1, qs, qe, fl);
    return fl == 0 ? min(t1, t2) : max(t1, t2);
  }
  void upd(int ss, int se, int si, int id, int k) {
    if (ss == se) {
      tree[si][0] = tree[si][1] = k;
      return;
    }
    if (id <= ((ss + se) >> 1)) {
      upd(ss, ((ss + se) >> 1), si * 2, id, k);
    } else {
      upd(((ss + se) >> 1) + 1, se, si * 2 + 1, id, k);
    }
    tree[si][0] = min(tree[si * 2][0], tree[si * 2 + 1][0]);
    tree[si][1] = max(tree[si * 2][1], tree[si * 2 + 1][1]);
  }
} seg[1 << K];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= k; ++j) {
      scanf("%d", &a[i][j]);
    }
  }
  for (int mask = 0; mask < (1 << k); mask++) {
    for (int i = 1; i <= n; i++) {
      int sum = 0;
      for (int j = 1; j <= k; j++) {
        if (((mask >> (j - 1)) & 1) == 0) {
          sum += a[i][j];
        } else {
          sum -= a[i][j];
        }
      }
      seg[mask].upd(1, n, 1, i, sum);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int i;
      scanf("%d", &i);
      for (int j = 1; j <= k; j++) {
        scanf("%d", &qt314[j]);
      }
      for (int mask = 0; mask < (1 << k); mask++) {
        int sum = 0;
        for (int j = 1; j <= k; j++) {
          if (((mask >> (j - 1)) & 1) == 0) {
            sum += qt314[j];
          } else {
            sum -= qt314[j];
          }
        }
        seg[mask].upd(1, n, 1, i, sum);
      }
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      int ans = -INF;
      for (int mask = 0; mask < (1 << k); ++mask) {
        int minv = seg[mask].query(1, n, 1, l, r, 0);
        int maxv = seg[mask].query(1, n, 1, l, r, 1);
        ans = max(ans, maxv - minv);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
