Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

1) 1+1+1+1+1+1+1 
2) 1+1+1+1+1+2 
3) 1+1+1+2+2 
4) 1+1+1+4 
5) 1+2+2+2 
6) 1+2+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 

Input

A single line with a single integer, N.

Output

The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).

Sample Input

7

Sample Output

6

當 i 為奇數時可以將 i-1 的展開項加 1 得到 i 的展開項
當 i 為偶數是單純的將 i-1 的展開項加 1 無法得到所有的 i 的展開項，因為 i 是偶數，所以 i 的展開項中有全為偶數的情況
將全為偶數的展開像除以 2 得到的是 i/2 的展開項（可以倒著想，將 i/2 的展開項乘 2 得到 i 的全偶數展開項） 

轉移式為 dp[i] = (i&1)?(dp[i-1]):(dp[i-1]+dp[i/2])

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4
 
 using namespace std;
 5 
 6 const int maxn = 1e6+5;
 7 int dp[maxn];
 8 
 9 int main(){
10     int n;
11     scanf("%d",&n);
12     memset(dp,0,sizeof(dp));
13     dp[0] = 1;
14 
15     for(int i=1;i<=n;++i){
16         if(i&1)
17             dp[i] = dp[i-1];
18         else{
19             dp[i] = dp[i-1
 
] + dp[i>>1];
20             if(dp[i]>1000000000)
21                 dp[i] -= 1000000000;
22         }
23     }
24 
25     printf("%d\n",dp[n]);
26     return 0;
27 }

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