Problem:
Score : 800 points

Problem Statement

There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.

It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.

If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)

All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.

Takahashi does not do shopping more than once in the same store.

Find the maximum number of times he can do shopping before time T + 0.5.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

0 \leq a_i \leq 10^9

0 \leq b_i \leq 10^9

0 \leq T \leq 10^9

Input

Input is given from Standard Input in the following format:

N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N

Output

Print the answer.

Sample Input 1

3 7
2 0
3 2
0 3

Sample Output 1

2

Here is one possible way to visit stores:

From time 0 to time 1: in 1 unit of time, he travels from his house to Store 1.

From time 1 to time 3: for 2 units of time, he stands in a queue for Store 1 to do shopping.

From time 3 to time 4: in 1 unit of time, he travels from Store 1 to Store 3.

From time 4 to time 7: for 3 units of time, he stands in a queue for Store 3 to do shopping.

In this way, he can do shopping twice before time 7.5.

Sample Input 2

1 3
0 3

Sample Output 2

0

Sample Input 3

5 21600
2 14
3 22
1 3
1 10
1 9

Sample Output 3

5

Sample Input 4

7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15

Sample Output 4

3

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

{$R+,S+,Q+,I+,O-}
{R-,S-,Q-,I-,O+}
// Code By H~$~C

uses math;

type
  int32 = longint;
  double = extended;
  bool = boolean;

const
  inf = 1061109567;
  lnf = 4557430888798830399;
  Maxn = 200005;
  LOG = 32;

var
  i, j, swap_tmp, x, y, z: int32;
  n, T, sza, szb, ans: int32;
  a: array [0 .. Maxn, 0 .. 1] of int32;
  b: array [0 .. Maxn] of int32;
  dp: array [0 .. LOG] of int32;

function qsort_less(i, j: int32): bool;
begin
  qsort_less := (a[j][0] * (a[i][1] + 1) < (a[i][0] * (a[j][1] + 1)));
end;
procedure qsort_swap(i, j: int32);
begin
  a[0] := a[i];
  a[i] := a[j];
  a[j] := a[0];
end;
procedure qsort(_l, _r: int32);
  var i, j, _p: int32;
begin
  if (_r = _l) then exit;
  i := _l; j := _r;
  _p := _l;
  repeat
    while qsort_less(i, _p) do inc(i);
    while qsort_less(_p, j) do dec(j);
    if (i <= j) then begin
      qsort_swap(i, j);
      inc(i); dec(j);
    end;
  until (i > j);
  if (i < _r) then qsort(i, _r);
  if (j > _l) then qsort(_l, j);
end;

function qsort2_less(i, j: int32): bool;
begin
  qsort2_less := (b[i] < b[j]);
end;
procedure qsort2_swap(i, j: int32);
begin
  b[0] := b[i];
  b[i] := b[j];
  b[j] := b[0];
end;
procedure qsort2(_l, _r: int32);
  var i, j, _p: int32;
begin
  if (_r = _l) then exit;
  i := _l; j := _r;
  _p := _l;
  repeat
    while qsort2_less(i, _p) do inc(i);
    while qsort2_less(_p, j) do dec(j);
    if (i <= j) then begin
      qsort2_swap(i, j);
      inc(i); dec(j);
    end;
  until (i > j);
  if (i < _r) then qsort2(i, _r);
  if (j > _l) then qsort2(_l, j);
end;

begin
//  assign(input,'in'); reset(input);
//  assign(output,'out'); rewrite(output);
  randomize;
  
  readln(n, T);
  sza := 0; szb := 0;
  for i := 1 to n do begin
    readln(x, y);
    if (x = 0) then begin
      inc(szb);
      b[szb] := y;
    end
    else begin
      inc(sza);
      a[sza, 0] := x;
      a[sza, 1] := y;
    end;
  end;
  
  for i := 2 to sza do begin
    x := random(i - 1) + 1;
    a[0] := a[x];
    a[x] := a[i];
    a[i] := a[0];
  end;
  qsort(1, sza);
  
  for i := 1 to LOG do dp[i] := T + 1;
  dp[0] := 0;
  for i := 1 to sza do begin
    for j := LOG - 1 downto 0 do begin
      if (dp[j] > T) then continue;
      if (dp[j + 1] > dp[j] + 1 + (dp[j] + 1) * a[i][0] + a[i][1]) then begin
        dp[j + 1] := dp[j] + 1 + (dp[j] + 1) * a[i][0] + a[i][1];
      end;
    end;
  end;
  
  for i := 2 to n do begin
    x := random(i - 1) + 1;
    b[0] := b[x];
    b[x] := b[i];
    b[i] := b[0];
  end;
  qsort2(1, szb);
  ans := 0;
  for i := 0 to LOG do begin
    if (dp[i] > T) then continue;
    x := T - dp[i];
    y := 0;
    for j := 1 to szb do begin
      if (x > b[j]) then begin
        dec(x, b[j] + 1);
        inc(y);
      end
      else break;
    end;
    if (i + y > ans) then ans := i + y;
  end;
  
  writeln(ans);
  
  close(input);
  close(output);
end.