### Problem Statement :

```We call an quadruple of positive integers, , beautiful if the following condition is true:

Note:  is the bitwise XOR operator.

Given , , , and , count the number of beautiful quadruples of the form  where the following constraints hold:

When you count the number of beautiful quadruples, you should consider two quadruples as same if the following are true:

They contain same integers.
Number of times each integers occur in the quadruple is same.
For example  and  should be considered as same.

nput Format

A single line with four space-separated integers describing the respective values of A, B, C, and D.```

### Solution :

```                            ```Solution in C :

In  C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int A;
int B;
int C;
int D;
long long int d[4097][3002][5];
long long int dp(int x,int i,int k){
if(k==0){
if(x!=0){
return 1;
}
return 0;
}
int p=0;
if(i<=A) p++;
if(i<=B) p++;
if(i<=C) p++;
if(i<=D) p++;
if(p==0) return 0;
int j=0;
if(p<k) return 0;
long long int l=0;
for(j=0;j<=k;j++){
if(d[x][i+1][k-j]==-1){
d[x][i+1][k-j] = dp(x, i+1, k-j);
}
l+=d[x][i+1][k-j];
x=x^i;
}
return l;
}

int main(){
memset(d, -1, sizeof(d));
scanf("%d %d %d %d",&A,&B,&C,&D);

printf("%lld\n",dp(0, 1, 4));
return 0;
}```
```

```                        ```Solution in C++ :

In  C++  :

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>

using namespace std;

#define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)
#define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)

int a[4];
int A,B,C,D;

typedef long long ll;

ll total;
ll cnt[10000];

int main(void){
int i,j;

REP(i,4) cin >> a[i];
sort(a, a+4);

A = a[0];
B = a[1];
C = a[2];
D = a[3];

for(i=1;i<=C;i++) for(j=i;j<=D;j++){
total++;
cnt[i^j]++;
}

ll ans = 0;

for(j=1;j<=B;j++){
for(i=1;i<=A&&i<=j;i++) ans += total - cnt[i^j];
for(i=j;i<=D;i++){
total--;
cnt[i^j]--;
}
}

cout << ans << endl;

return 0;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int A = in.nextInt();
int B = in.nextInt();
int C = in.nextInt();
int D = in.nextInt();

int[] m = new int[] {A, B, C, D};
Arrays.sort(m);

long[][] count = new long[3001][4096];

for (int i=1; i<=m[0]; i++) {
for (int j=i; j<=m[1]; j++) {
count[j][i ^ j] ++;
}
}
for (int i=0; i<4096; i++) {
for (int j=1; j<=3000; j++) {
count[j][i] += count[j-1][i];
}
}

long[] sum = new long[3001];
for (int j=1; j<=3000; j++) {
for (int i=0; i<4096; i++) {
sum[j] += count[j][i];
}
}

long res=0;
for (int i=1; i<=m[2]; i++) {
for (int j=i; j<=m[3]; j++) {
int x = i ^ j;
res += sum[i] - count[i][i^j];
}
}
System.out.println(res);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

#!/bin/python3

import sys

A,B,C,D = input().strip().split(' ')
temp = [int(A),int(B),int(C),int(D)]
temp.sort()
A,B,C,D=temp

tot=[0]*4100

tree=[([0]*3100).copy() for i in range(4100)]
cnt=[([0]*3100).copy() for i in range(4100)]
def updateBIT(treeNo,ind):
while(ind<3100):
tree[treeNo][ind]+=1
ind+=ind&-ind

def getBITVal(treeNo,ind):
ret=0
while(ind>0):
ret+=tree[treeNo][ind]
ind-=ind&-ind
return ret

zero=0
tot=[0]*3100
for i in range(1,A+1):
for j in range(i,B+1):
#updateBIT(i^j,j)
cnt[i^j][j]+=1
tot[j]+=1

for i in range(len(cnt)):
for j in range(1,len(cnt[i])):
cnt[i][j]+=cnt[i][j-1]

for i in range(1,len(tot)):
tot[i]+=tot[i-1]

ans=0
for i in range(1,C+1):
ans+=tot[i]*(D+1-i)
for j in range(i,D+1):
zero+=cnt[i^j][i]
#ans+=getBITVal2(i)
#print(i,j)
#print(ans)
ans-=zero
#print(zero)
print(ans)```
```

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