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### Heads I Win, Tails You Lose

Difficulty: ★★★☆☆

##### Problem statement

The rivalry between Michael and Ray seems endless. These two boys even compete with each other by coin flipping.

Just now, Michael suggested a game with the following rules:

“Each person will flip a separate coin. If it lands on heads, the person who flipped the coin gets 1 point, otherwise no point is granted. After n flips, the person who has more points wins.”

Being fed up with such childish competition between his two friends, David comes to you in a need for support so that he can code the probability of Michael winning over Ray.

Input

– Enter n denoting the number of flips.

Output

– Print the number of cases where Michael wins over Ray mod 10^9+7, as the number can be very large.

Constraints:

– n <= 20

Sample Input

2

Sample Output

5

##### Answer

Suppose the number of times that one player has k point is cnt[k].

Then the result will be: sum of (cnt[k] * sum of (cnt[i]) with i < k).

We can use bitmask to compute cnt[k].

Here is sample code.

`#include <iostream>`

#include <string.h>

using namespace std;

#define MOD 1000000007

const int N = 21;

int n;

unsigned long long cnt[N];

int main(){

ios_base::sync_with_stdio(false);

cin >> n;

memset(cnt, 0, sizeof(cnt));

for(int i = 0; i < (1 << n); ++i) {

cnt[__builtin_popcount(i)]++;

}

int sum = 0, ans = 0;

for(int i = 0; i <= n; ++i){

ans = (ans + sum * cnt[i]) % MOD;

sum = (sum + cnt[i]) % MOD;

}

cout << ans << endl;

return 0;

}