# Codeforces 914 C. Travelling Salesman and Special Numbers （dp）

## Description

The Travelling Salesman spends a lot of time travelling so he tends to get bored. To pass time, he likes to perform operations on numbers. One such operation is to take a positive integer x and reduce it to the number of bits set to 1 in the binary representation of x. For example for number 13 it’s true that $13_{10} = 1101_2$ , so it has 3 bits set and 13 will be reduced to 3 in one operation.

He calls a number special if the minimum number of operations to reduce it to 1 is k.

He wants to find out how many special numbers exist which are not greater than n. Please help the Travelling Salesman, as he is about to reach his destination!

Since the answer can be large, output it modulo 10^9 + 7.

## Input

The first line contains integer n (1 ≤ n < 2^1000).

The second line contains integer k (0 ≤ k ≤ 1000).

Note that n is given in its binary representation without any leading zeros.

## Output

Output a single integer — the number of special numbers not greater than n, modulo 10^9 + 7.

## Examples input

110
2


## Examples output

3


## AC 代码

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
const int maxn = 1e5+10;

char str[maxn];
LL mul[maxn];
LL inv[maxn];
int dp[maxn];
int len,k;

void init()
{
mul[0]=1;
for(int i=1; i<maxn; i++)
mul[i]=(mul[i-1]*i)%mod;

inv[0]=inv[1]=1;
for(int i=2; i<maxn; i++)
inv[i]=(LL)(mod-mod/i)*inv[mod%i]%mod;
for(int i=1; i<maxn; i++)
inv[i]=(inv[i-1]*inv[i])%mod;
}

LL C(int n,int m)
{
return mul[n]*inv[m]%mod*inv[n-m]%mod;
}

int getNum(int x)
{
int res = 0;
while(x)
{
res+=x&1;
x>>=1;
}
return res;
}

int call(int x)
{
int res = 0;
for(int i=0; i<len; i++)
if(str[i]-'0')
res = (res + C(len-i-1,x--))%mod;
return (res + (x==0))%mod;
}

void solve()
{
len = strlen(str);
if(k==0)
cout<<1<<endl;
else if(k==1)
cout<<len-1<<endl;
else
{
int ans = 0;
for(int i=2; i<=1000; i++)
{
dp[i] = dp[getNum(i)] + 1;
if(dp[i]==k-1)
ans = (ans + call(i)) % mod;
}
cout<<ans<<endl;
}
}

int main()
{
IO;
init();
cin>>str>>k;
solve();
return 0;
}