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;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int n, k, a[5], q;
struct seg {
  int l, r, maxn[32], minn[32];
} t[N << 2];
inline void cmax(int &x, int y) {
  if (x < y) x = y;
}
inline void cmin(int &x, int y) {
  if (x > y) x = y;
}
void pushup(int p) {
  for (int i = 0; i < (1 << k); i++) {
    t[p].maxn[i] = max(t[p << 1].maxn[i], t[p << 1 | 1].maxn[i]);
    t[p].minn[i] = min(t[p << 1].minn[i], t[p << 1 | 1].minn[i]);
  }
}
void build(int p, int l, int r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    for (int i = 0; i < k; i++) cin >> a[i];
    for (int i = 0; i < (1 << k); i++) {
      int sum = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1)
          sum += a[j];
        else
          sum -= a[j];
      }
      t[p].maxn[i] = t[p].minn[i] = sum;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  pushup(p);
}
void upd(int p, int idx) {
  int l = t[p].l, r = t[p].r;
  if (l == r) {
    for (int i = 0; i < k; i++) cin >> a[i];
    for (int i = 0; i < (1 << k); i++) {
      int sum = 0;
      for (int j = 0; j < k; j++) {
        if ((i >> j) & 1)
          sum += a[j];
        else
          sum -= a[j];
      }
      t[p].maxn[i] = t[p].minn[i] = sum;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (idx <= mid)
    upd(p << 1, idx);
  else
    upd(p << 1 | 1, idx);
  pushup(p);
}
int askmax(int p, int x, int y, int idx) {
  int l = t[p].l, r = t[p].r;
  if (l >= x && r <= y) return t[p].maxn[idx];
  int mid = (l + r) >> 1, ans = -INF;
  if (x <= mid) cmax(ans, askmax(p << 1, x, y, idx));
  if (y > mid) cmax(ans, askmax(p << 1 | 1, x, y, idx));
  return ans;
}
int askmin(int p, int x, int y, int idx) {
  int l = t[p].l, r = t[p].r;
  if (l >= x && r <= y) return t[p].minn[idx];
  int mid = (l + r) >> 1, ans = INF;
  if (x <= mid) cmin(ans, askmin(p << 1, x, y, idx));
  if (y > mid) cmin(ans, askmin(p << 1 | 1, x, y, idx));
  return ans;
}
int solve(int l, int r) {
  int res = 0;
  for (int i = 0; i < (1 << k); i++)
    cmax(res, askmax(1, l, r, i) - askmin(1, l, r, i));
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  build(1, 1, n);
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x;
      cin >> x;
      upd(1, x);
    } else {
      int l, r;
      cin >> l >> r;
      cout << solve(l, r) << '\n';
    }
  }
  return 0;
}
