Problem: 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.

Examples

Input

3 7
2 0
3 2
0 3


Output

2


Input

1 3
0 3


Output

0


Input

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


Output

5


Input

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


Output

3

Solution (in PYTHON3):

Solution:
def examA():
    S = SI()
    if len(S)%2==1:
        print("No")
        return
    for i in range(len(S)//2):
        if S[i*2:i*2+2]!="hi":
            print("No")
            return
    print("Yes")
    return

def examB():
    a, b, M = LI()
    A = LI()
    B = LI()
    D = [LI()for _ in range(M)]
    ans = min(A) + min(B)
    for x,y,c in D:
        cur = A[x-1]+B[y-1]-c
        ans = min(ans,cur)
    print(ans)
    return

def examC():
    def bfs(n, e, fordfs):
        # 点の数、スタートの点、有向グラフ
        W = [-1] * n
        # 各点の状態量、最短距離とか,見たかどうかとか
        W[e] = 0
        que = deque()
        que.append(e)
        while que:
            now = que.popleft()
            nowW = W[now]
            for ne in fordfs[now]:
                if W[ne] == -1:
                    W[ne] = nowW + 1
                    que.append(ne)
        return W
    N = I()
    V = [[]for _ in range(N)]
    for _ in range(N-1):
        a, b = LI()
        a -= 1
        b -= 1
        V[a].append(b)
        V[b].append(a)
    L = bfs(N,0,V)
    #print(L)
    odd = 0
    for l in L:
        if l==-1:
            print(-1)
            return
        if l%2==1:
            odd += 1
    G = []
    if odd*2<N:
        for i in range(N):
            if L[i]%2==1:
                G.append(i)
    else:
        for i in range(N):
            if L[i]%2==0:
                G.append(i)

    used = [False]*(N+1)
    ans = [0]*N
    if N//3>=len(G):
        cur = 3
        for g in G:
            ans[g] = cur
            used[cur] = True
            cur += 3
        cur = 1
        for i in range(N):
            if ans[i]!=0:
                continue
            while(used[cur]):
                cur += 1
            ans[i] = cur
            used[cur] = True
    else:
        cur = 1
        for g in G:
            ans[g] = cur
            used[cur] = True
            cur += 3
            if cur>N:
                cur = 3
        cur = 1
        for i in range(N):
            if ans[i]!=0:
                continue
            while(used[cur]):
                cur += 1
            ans[i] = cur
            used[cur] = True
    print(" ".join(map(str,ans)))
    return

def examD():
    N, T = LI()
    A = []
    A0 = []
    for _ in range(N):
        a, b = LI()
        if a==0:
            A0.append(b)
        else:
            A.append((a/(b+1),a,b))
    A.sort(reverse=True)
    n = len(A)
    S = min(31,n)
    dp = [[inf]*(n+1) for _ in range(S+1)]
    dp[0][0] = 0
    for i,(_,a,b) in enumerate(A):
        for j in range(S):
            now = dp[j][i] + 1
            dp[j][i+1] = min(dp[j][i+1],dp[j][i])
            if now==inf:
                continue
            dp[j+1][i+1] = min(dp[j+1][i+1],now+a*now+b)
    #print(dp)
    dp0 = [0]*(len(A0)+1)
    A0.sort()
    for i,b in enumerate(A0):
        dp0[i+1] = dp0[i]+b+1
    ans = 0
    for i in range(S+1):
        t = T - min(dp[i])
        if t<0:
            break
        cur = (bisect.bisect_right(dp0,t)-1) + i
        if ans<cur:
            ans = cur
    print(ans)
    return

def examE():
    N, M = LI()
    H = 2**N-1; W = 2**M -1
    ans = [[0]*W for _ in range(H)]
    for h in range(H):
        flagh = False
        for m in range(1,11)[::-1]:
            if (h+1)%(2**m)==0:
                flagh = 2**m*2
        for w in range(W):
            flagw = False
            for m in range(1,11)[::-1]:
                if (w+1)%(2**m)==0:
                    flagw = 2 ** m * 2
            if flagh and flagw:
                continue
            ans[h][w] = 1
    for v in ans:
        print("".join(map(str,v)))
    return

def examF():
    ans = 0
    print(ans)
    return

import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**6)

if __name__ == '__main__':
    examD()

"""

"""