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

1310=11012${13}_{10}={1101}_2$ , 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

題意

我們定義一種操作為將一個正整數變化為它二進位制 1 的個數，問在小於等於給定二進位制數的正整數中，有多少個數可以經過 k 次操作變化為 1 。

思路

從題中我們可以看到，一次操作可以將一個數變化為其二進位制 1 的個數，於是我們設 dp[i] 表示二進位制中有 i 個 1 的數需要幾次變化才能到達 1 。

顯然 dp[i] = dp[getNum(i)] + 1 ，其中 getNum 可以得出 i 二進位制 1 的個數。

對於 dp[i] == k-1 的 i ，接下來的工作便是找出小於等於給定數且二進位制 1 的個數為 i 的數的數量，直接用組合數很方便可以求得。

AC 程式碼

#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;
}