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;
int main() {}
const int MAXN = 2e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long MOD = 1e9 + 7;
const double PI = acos(-1.0);
struct Node {
  int Max[32], Min[32];
} tree[MAXN << 2];
int b[MAXN][5];
int n, K;
void pushUp(int rt) {
  for (int i = 0; i < (1 << K); ++i) {
    tree[rt].Max[i] = max(tree[rt << 1].Max[i], tree[rt << 1 | 1].Max[i]);
    tree[rt].Min[i] = min(tree[rt << 1].Min[i], tree[rt << 1 | 1].Min[i]);
  }
}
void build(int l, int r, int rt) {
  if (l == r) {
    for (int i = 0; i < (1 << K); ++i) {
      tree[rt].Max[i] = 0;
      tree[rt].Min[i] = 0;
      for (int j = 0; j < K; ++j) {
        if ((i >> j) & 1) {
          tree[rt].Max[i] += b[l][j];
          tree[rt].Min[i] += b[l][j];
        } else {
          tree[rt].Max[i] += -b[l][j];
          tree[rt].Min[i] += -b[l][j];
        }
      }
    }
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushUp(rt);
}
void update(int l, int r, int p, int rt) {
  if (l == r && l == p) {
    for (int i = 0; i < (1 << K); ++i) {
      tree[rt].Max[i] = 0;
      tree[rt].Min[i] = 0;
      for (int j = 0; j < K; ++j) {
        if ((i >> j) & 1) {
          tree[rt].Max[i] += b[l][j];
          tree[rt].Min[i] += b[l][j];
        } else {
          tree[rt].Max[i] += -b[l][j];
          tree[rt].Min[i] += -b[l][j];
        }
      }
    }
    return;
  }
  int mid = (l + r) / 2;
  if (p <= mid) {
    update(l, mid, p, rt << 1);
  } else {
    update(mid + 1, r, p, rt << 1 | 1);
  }
  pushUp(rt);
}
int query1(int L, int R, int l, int r, int rt, int mask) {
  if (L <= l && r <= R) {
    return tree[rt].Max[mask];
  }
  int mid = (l + r) / 2;
  int ans = -1000000000;
  if (L <= mid) {
    ans = max(ans, query1(L, R, l, mid, rt << 1, mask));
  }
  if (R > mid) {
    ans = max(ans, query1(L, R, mid + 1, r, rt << 1 | 1, mask));
  }
  return ans;
}
int query2(int L, int R, int l, int r, int rt, int mask) {
  if (L <= l && r <= R) {
    return tree[rt].Min[mask];
  }
  int mid = (l + r) / 2;
  int ans = 1000000000;
  if (L <= mid) {
    ans = min(ans, query2(L, R, l, mid, rt << 1, mask));
  }
  if (R > mid) {
    ans = min(ans, query2(L, R, mid + 1, r, rt << 1 | 1, mask));
  }
  return ans;
}
int hzkFather() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < K; ++j) {
      scanf("%d", &b[i][j]);
    }
  }
  build(1, n, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int i;
      scanf("%d", &i);
      for (int j = 0; j < K; ++j) scanf("%d", &b[i][j]);
      update(1, n, i, 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = -1000000000;
      for (int i = 0; i < (1 << K); ++i) {
        int tmp1 = query1(l, r, 1, n, 1, i);
        int tmp2 = query2(l, r, 1, n, 1, i);
        ans = max(ans, tmp1 - tmp2);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
int neverUsed = hzkFather();
