B. Dima and Sequence

Dima got into number sequences. Now he's got sequence?a1,?a2,?...,?an, consisting of?n?positive integers. Also, Dima has got a function?f(x), which can be defined with the following recurrence:

• f(0)?=?0;
• f(2·x)?=?f(x);
• f(2·x?+?1)?=?f(x)?+?1.

Dima wonders, how many pairs of indexes?(i,?j)?(1?≤?i?<?j?≤?n)?are there, such that?f(ai)?=?f(aj). Help him, count the number of such pairs.

Input

The first line contains integer?n?(1?≤?n?≤?105). The second line contains?n?positive integers?a1,?a2,?...,?an?(1?≤?ai?≤?109).

The numbers in the lines are separated by single spaces.

Output

In a single line print the answer to the problem.

Please, don't use the?%lld?specifier to read or write 64-bit integers in?C++. It is preferred to use the?cin,?cout?streams or the?%I64dspecifier.

Examples

input

3 1 2 4

output

3

input

3 5 3 1

output

1

Note

In the first sample any pair?(i,?j)?will do, so the answer is?3.

In the second sample only pair?(1,?2)?will do.

打表100條，找到規(guī)律。

偶數(shù)找規(guī)律，計(jì)數(shù)找祖宗。

#include<iostream> #include<queue> #include<algorithm> #include<set> #include<cmath> #include<vector> #include<map> #include<stack> #include<bitset> #include<cstdio> #include<cstring> //---------------------------------Sexy operation--------------------------//#define cini(n) scanf("%d",&n) #define cinl(n) scanf("%lld",&n) #define cinc(n) scanf("%c",&n) #define cins(s) scanf("%s",s) #define coui(n) printf("%d",n) #define couc(n) printf("%c",n) #define coul(n) printf("%lld",n) #define debug(n) printf("%d_________________________________\n",n); #define speed ios_base::sync_with_stdio(0) #define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout) //-------------------------------Actual option------------------------------// #define rep(i,a,n) for(int i=a;i<=n;i++) #define per(i,n,a) for(int i=n;i>=a;i--) #define Swap(a,b) a^=b^=a^=b #define Max(a,b) (a>b?a:b) #define Min(a,b) a<b?a:b #define mem(n,x) memset(n,x,sizeof(n)) #define mp(a,b) make_pair(a,b) #define pb(n) push_back(n) //--------------------------------constant----------------------------------//#define INF 0x3f3f3f3f #define maxn 10000000 #define esp 1e-9 using namespace std; typedef pair<int,int>PII; typedef pair<string,int>PSI; typedef long long ll; //___________________________Dividing Line__________________________________// long long a[40]= {0}; int main() {int n,x;cini(n);while(n--){int sum=0;cini(x);while(x){if(x&1){sum++;x=(x-1)/2;}else x/=2;}a[sum]++;}long long ans=0;for(int i=0; i<40; i++)ans+=a[i]*(a[i]-1)/2;cout<<ans<<endl;return 0; }

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