Problem: While resting on the ship after the "Russian Code Cup" a boy named Misha invented an interesting game. He promised to give his quadrocopter to whoever will be the first one to make a rectangular table of size n × m, consisting of positive integers such that the sum of the squares of numbers for each row and each column was also a square.

Since checking the correctness of the table manually is difficult, Misha asks you to make each number in the table to not exceed 108.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the table. 

Output

Print the table that meets the condition: n lines containing m integers, separated by spaces. If there are multiple possible answers, you are allowed to print anyone. It is guaranteed that there exists at least one correct answer.

Examples

Input

1 1


Output

1

Input

1 2


Output

3 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:32000000")
using namespace std;
const int INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
int n, m;
int a[10][10];
bool issq(long long x) {
  int sq = (int)(sqrt(x + .0) + .5);
  return (long long)sq * sq == x;
}
bool check() {
  bool ok = true;
  for (int i = 0; i < n; ++i) {
    long long v1 = 0;
    for (int j = 0; j < m; ++j) {
      v1 += a[i][j] * (long long)a[i][j];
    }
    ok &= issq(v1);
    if (!ok) return ok;
  }
  for (int j = 0; j < m; ++j) {
    long long v1 = 0;
    for (int i = 0; i < n; ++i) {
      v1 += a[i][j] * (long long)a[i][j];
    }
    ok &= issq(v1);
    if (!ok) return ok;
  }
  return ok;
}
void rec(int i, int j) {
  if (i == n) {
    if (check()) {
      for (int p1 = 0; p1 < n; ++p1) {
        for (int p2 = 0; p2 < m; ++p2) printf("%d ", a[p1][p2]);
        printf("\n");
      }
      printf("\n");
    }
    return;
  }
  int ni = i;
  int nj = j + 1;
  if (nj == m) nj = 0, ni++;
  for (int iter = 0; iter < 9; ++iter) {
    a[i][j] = iter + 1;
    rec(ni, nj);
  }
}
vector<int> solve(int n) {
  if (n == 2) {
    vector<int> r;
    r.push_back(3);
    r.push_back(4);
    return r;
  }
  if (n % 2 == 0) {
    vector<int> r;
    for (int i = 0; i < n - 1; ++i) r.push_back(1);
    r.push_back((n - 2) / 2);
    return r;
  } else {
    vector<int> r;
    for (int i = 0; i < n - 2; ++i) r.push_back(1);
    r.push_back(2);
    r.push_back((n + 1) / 2);
    return r;
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> v1 = solve(n);
  vector<int> v2 = solve(m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      int res = v1[i] * v2[j];
      printf("%d ", res);
    }
    printf("\n");
  }
  return 0;
}