#include <bits/stdc++.h>
#define maxn 100
#define pii pair<int,int>
#define fi first
#define se second
#define inf INT_MAX/2-1

using namespace std;

int v[2][maxn+5], mdp[maxn+5][maxn+5];
int dp[maxn+5][maxn+5]; /// dp[i][j] -- nr maxim de litri B care pot fi bauti in timpul ramas din cele tim secunde cu exact i litri bauti din
/// tipul a numai cu oamenii din [1,j]
pii prew[maxn+5][maxn+5], mprew[maxn+5][maxn+5];
int n, l;

ifstream fin ( "lapte.in" );
ofstream fout ( "lapte.out" );

void mem ( int n, int l )
{
  for ( int i = 0; i <= l; i++ )
    for ( int j = 0; j <= n; j++ ) mdp[i][j] = dp[i][j], mprew[i][j] = prew[i][j];
}

bool _try ( int tim )
{
  int i, j, k, z, _m;
  for ( i = 0; i <= l; i++ )
    for ( j = 0; j <= n; j++ ) dp[i][j] = 0;
  for ( i = 1; i <= l; i++ ) dp[i][1] = (tim - i*v[0][1]) / v[1][1];
  for ( j = 2; j <= n; j++ ) dp[1][j] = max(dp[1][j-1], (tim-v[0][j])/v[1][j]);

  for ( i = 1; i <= l; i++ )
    for ( j = 2; j <= n; j++ )
      for ( k = 1; k <= i; k++ )
        if ( dp[i][j] < dp[i-k][j-1] + (tim-k*v[0][j])/v[1][j] ) dp[i][j] = dp[i-k][j-1] + (tim-k*v[0][j])/v[1][j], prew[i][j] = {i,k};
  return (dp[l][n] >= l);
}

void sol ( int i, int j )
{
  int ni = mprew[i][j].fi - mprew[i][j].se, nj = j-1;
  if ( nj >= 1 ) sol (ni,nj);
  fout << mprew[i][j].se << ' ' << mdp[i][j] - mdp[ni][nj] << '\n';
}

int main ()
{
  fin >> n >> l;

  int i, j, k, z;
  for ( i = 1; i <= n; i++ ) fin >> v[0][i] >> v[1][i];

  int pas = (1<<7), tim = (1<<7);
  for ( ; pas > 0; pas /= 2 )
    if ( tim - pas >= 0 && _try(tim-pas) == 1 ) tim -= pas, mem(n,l);

  fout << tim << '\n';
  sol (l,n);

  fin.close ();
  fout.close ();

  return 0;
}