Sasha and a Bit of Relax（前缀异或）

Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.

Therefore, Sasha decided to upsolve the following problem:

You have an array aa with nn integers. You need to count the number of funny pairs (l,r)(l,r) (l≤r)(l≤r). To check if a pair (l,r)(l,r) is a funny pair, take mid=l+r−12mid=l+r−12, then if r−l+1r−l+1 is an even number and al⊕al+1⊕…⊕amid=amid+1⊕amid+2⊕…⊕aral⊕al+1⊕…⊕amid=amid+1⊕amid+2⊕…⊕ar, then the pair is funny. In other words, ⊕⊕ of elements of the left half of the subarray from ll to rr should be equal to ⊕⊕ of elements of the right half. Note that ⊕⊕ denotes the bitwise XOR operation.

It is time to continue solving the contest, so Sasha asked you to solve this task.

Input

The first line contains one integer nn (2≤n≤3⋅1052≤n≤3⋅105) — the size of the array.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai<2200≤ai<220) — array itself.

Output

Print one integer — the number of funny pairs. You should consider only pairs where r−l+1r−l+1 is even number.

Examples

Input

5
1 2 3 4 5

Output

1

Input

6
3 2 2 3 7 6

Output

3

Input

3
42 4 2

Output

0

Note

Be as cool as Sasha, upsolve problems!

In the first example, the only funny pair is (2,5)(2,5), as 2⊕3=4⊕5=12⊕3=4⊕5=1.

In the second example, funny pairs are (2,3)(2,3), (1,4)(1,4), and (3,6)(3,6).

In the third example, there are no funny pairs.

给你一个数列 求有多少区间满足

长度为偶数&&前一部分异或值等于后一部分异或值

用代码模拟一下标红样例即可

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <queue>
#include <map>
#define sf(a) scanf("%d\n",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define E 1e-8
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int idata=1e5+5;

int n,m,t;
int flag,judge,temp;
ll ans,cnt,sum,len;
int minn,maxx;
//int step[idata];
int i,j,k;
int a[1<<20+3][2];

int main()
{
    while(cin>>n)
    {
        a[0][0]=1;
        //前缀为0，且为偶数
        for(i=1;i<=n;i++)
        {
            cin>>m;
            sum^=m;
            ans+=a[sum][i&1]++;
        }
        cout<<ans<<endl;
    }
    return 0;
}