C. Amr and Chemistry

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.

Amr has n different types of chemicals. Each chemical i has an initial volume of ai liters. For this experiment, Amr has to mix all the chemicals together, but all the chemicals volumes must be equal first. So his task is to make all the chemicals volumes equal.

To do this, Amr can do two different kind of operations.

• Choose some chemical i and double its current volume so the new volume will be 2ai
• Choose some chemical i and divide its volume by two (integer division) so the new volume will be 

Suppose that each chemical is contained in a vessel of infinite volume. Now Amr wonders what is the minimum number of operations required to make all the chemicals volumes equal?

Input

The first line contains one number n (1 ≤ n ≤ 105), the number of chemicals.

The second line contains n space separated integers ai (1 ≤ ai ≤ 105), representing the initial volume of the i-th chemical in liters.

Output

Output one integer the minimum number of operations required to make all the chemicals volumes equal.

Sample test(s)

input

34 8 2

output

2

input

33 5 6

output

5

Note

In the first sample test, the optimal solution is to divide the second chemical volume by two, and multiply the third chemical volume by two to make all the volumes equal 4.

In the second sample test, the optimal solution is to divide the first chemical volume by two, and divide the second and the third chemical volumes by two twice to make all the volumes equal 1.

将每一个数写为2进制。计算公共前缀

答案必为 [最长公共前缀(LCS)]+00...00

证：

若LCS后有1

则每一个数都要在对应位置有1，

故LCS可增长。矛盾

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              using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i
              
               =0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (100000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n,a[MAXN],a2[MAXN],h[MAXN]={0},q[MAXN]={0};int main(){ freopen("C.in","r",stdin);// freopen(".out","w",stdout); cin>>n; For(i,n) scanf("%d",&a[i]); sort(a+1,a+1+n); memcpy(a2,a,sizeof(a)); int p=a[1]; For(i,n) { while (p^a[i]) { if (p
               
                >=1; else p>>=1; } } memcpy(a,a2,sizeof(a)); For(i,n) { while (a[i]^p) { h[i]--; if (a[i]&1) q[i]=h[i]; a[i]>>=1; } } ll ans=INF; Rep(t,100) { ll p=0; For(i,n) { if (q[i]==0||h[i]+t<=q[i]) p+=abs(h[i]+t); else p+=abs(q[i])+abs(q[i]-(h[i]+t)); } ans=min(ans,p); } cout<
                
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