Rest (XOR) position and thinking

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So if someone contributes , There must be someone in every paragraph 1, So we can consider from high to low , then b[j]=arr[j]&p Reverse operation Finally, judge whether the values are equal , If the number is even, it can be XOR cancelled

#include<bits/stdc++.h>
using namespace std;
const int N=3e5+9;
int arr[N],b[N];
int main(){
    
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)cin>>arr[i];
    int p=0;
    for(int i=30;i>=0;i--){
    
         p+=(1<<i);
         for(int j=1;j<=n;j++)b[j]=arr[j]&p;// First & operation 
         int sum=0,d=0;
        for(int j=1;j<=n;j++){
    
            sum^=b[j];// Or else 
            if(sum==p){
    
                d++;
                sum=0;
            }
        }
        if(sum==0&&d>=m&&(d-m)%2==0);// If the segment greater than is even, the XOR is 0 Equivalent 
        else p-=(1<<i);
    }
    cout<<p<<endl;
    return 0;
}