Problem: People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.

Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n × n is called k-special if the following three conditions are satisfied:

  * every integer from 1 to n2 appears in the table exactly once; 
  * in each row numbers are situated in increasing order; 
  * the sum of numbers in the k-th column is maximum possible. 



Your goal is to help Alice and find at least one k-special table of size n × n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 500, 1 ≤ k ≤ n) — the size of the table Alice is looking for and the column that should have maximum possible sum.

Output

First print the sum of the integers in the k-th column of the required table.

Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.

If there are multiple suitable table, you are allowed to print any.

Examples

Input

4 1


Output

28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16


Input

5 3


Output

85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void c_p_c() {}
class DisjointSet {
  unordered_map<int, int> parent;

 public:
  void makeSet(int N) {
    for (int i = 0; i < N; i++) parent[i] = i;
  }
  int Find(int v) {
    if (v == parent[v]) return v;
    return parent[v] = Find(parent[v]);
  }
  void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    parent[x] = y;
  }
};
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void PrimeFactors(long long n) {
  while (n % 2 == 0) {
    cout << 2 << " ";
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      cout << i << " ";
      n = n / i;
    }
  }
  if (n > 2) cout << n << " ";
}
template <typename T>
void parray(T a[], T n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
vector<long long> uniqueelement(vector<long long> &a) {
  auto ip = unique((a).begin(), (a).end());
  a.resize(distance(a.begin(), ip));
  return a;
}
bool IsPrime(long long x) {
  if (x == 1) return false;
  for (int i = 2; 1ll * i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void leftrotate(string &s, int d) {
  reverse(s.begin(), s.begin() + d);
  reverse(s.begin() + d, s.end());
  reverse(s.begin(), s.end());
}
void rightrotate(string &s, int d) { leftrotate(s, s.length() - d); }
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
vector<long long> sieve() {
  vector<long long> primes;
  bool a[1000000] = {0};
  int mprimes = 1000000;
  a[0] = a[1] = true;
  for (int i = 2; i * i < mprimes; ++i) {
    if (a[i] == false) {
      for (int j = i * i; j < mprimes; j = j + i) {
        a[j] = true;
      }
    }
  }
  for (int i = 2; i < mprimes; ++i) {
    if (!a[i]) primes.push_back(i);
    ;
  }
  return primes;
}
void Ynot() {
  long long i, n, m, k, c = 0, sum = 0;
  bool flag = false;
  cin >> n >> k;
  int a[n + 1][n + 1];
  int val = 1;
  for (int i = 1; i < n + 1; ++i) {
    for (int j = 1; j < k; ++j) {
      a[i][j] = val++;
    }
  }
  for (int i = 1; i < n + 1; ++i) {
    for (int j = k; j < n + 1; ++j) {
      a[i][j] = val++;
      if (j == k) sum += a[i][j];
    }
  }
  cout << sum << "\n";
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      cout << a[i][j] << " ";
    }
    cout << "\n";
  }
}
int main() {
  c_p_c();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    Ynot();
  }
  return 0;
}