#include <bits/stdc++.h>
#define int long long
using namespace std;
#define ll long long
vector<vector<ll>> solve(ll n,ll k,vector<ll>&a,vector<ll>&b)
{
    priority_queue<pair<double,pair<ll,ll>>> pq;
    for(int i=0;i<n;i++)
    {
        double x=a[i];
        double y=b[i];
        double dis=sqrt(x+y);
        pq.push({dis,{x,y}});
        if(pq.size()>k)  pq.pop();
    }
    vector<vector<ll>>ans;
    while(!pq.empty())
    {
        ans.push_back({pq.top().second.first,pq.top().second.second});
        pq.pop();
    }
    sort(begin(ans),end(ans));
    return ans;
}
signed main()
{       
        ll n,k; cin>>n>>k;
        vector<ll>a(n),b(n);
        for(ll i=0;i<n;i++) cin>>a[i];
        for(ll i=0;i<n;i++) cin>>b[i];
        vector<vector<ll>>ans=solve(n,k,a,b);
        for(auto it:ans) cout<<it[0]<<" "<<it[1]<<endl;
           
    return 0;
}


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def is_palindrome(s):
    return s == s[::-1]

def longest_palindrome_from_substrings(A):
    palindromes = []
    pairs = []
    max_single_palindrome = ""

    for s in A:
        if is_palindrome(s):
            palindromes.append(s)
            if len(s) > len(max_single_palindrome):
                max_single_palindrome = s

    for i in range(len(A)):
        for j in range(i + 1, len(A)):
            combined1 = A[i] + A[j]
            combined2 = A[j] + A[i]
            if is_palindrome(combined1):
                pairs.append(combined1)
            if is_palindrome(combined2):
                pairs.append(combined2)

    longest_palindrome = max_single_palindrome
    for p in pairs:
        if len(p) > len(longest_palindrome):
            longest_palindrome = p

    return longest_palindrome



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def split_string_cost(S):
    # Length of the string S
    len_S = len(S)
   
    # To store the cost of the split parts
    max_cost = 0
   
    # Set to keep track of distinct characters in the first part
    distinct_chars_A = set()
   
    # List to keep track of the cost for the second part from each split position
    cost_B = [0] * len_S
   
    # Set to keep track of distinct characters in the second part
    distinct_chars_B = set()
   
    # Calculate cost for second part from the end
    for i in range(len_S - 1, -1, -1):
        distinct_chars_B.add(S[i])
        cost_B[i] = len(distinct_chars_B)
   
    # Calculate maximum sum of cost for parts A and B
    for i in range(len_S - 1):
        distinct_chars_A.add(S[i])
        cost_A = len(distinct_chars_A)
        cost = cost_A + cost_B[i + 1]
        max_cost = max(max_cost, cost)
   
    # Calculate the result as |S| - X
    result = len_S - max_cost
    return result

# Example usage
S = "aaabbb"
print(split_string_cost(S))  # Output: 3

#include <bits/stdc++.h>
using namespace std;

int equalzeroandone(vector<int>v){
    int n=v.size();
    for(int i=0;i<n;i++){
        if(v[i]==0){
            v[i]=-1;
        }
    }
    int sum=0;
    int ans=-1;
   map<int,int>mp;
    for(int i=0;i<n;i++){
        sum+=v[i];
        if(sum==0){
            ans=i+1;
        }
        if(mp.find(sum)!=mp.end()){
            ans=max(ans,i-mp[sum]);
        }
        else{
            mp[sum]=i;
        }
    }
    return ans;
}

int main() {
    int n;
    cin>>n;
    vector<int>v(n);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    cout<<equalzeroandone(v);
}

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