观察找规律
最大的和二分最小的，然后交替相邻，看是否合法
找到最长的合法段
答案就是总长度减去合法段加上原先的长度
#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<" "<<#e<<" = "<<e<<endl;
#define debug0(x) cout << "debug0: " << x << endl
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define YES cout<<"Yes"<<endl
#define NO cout<<"No"<<endl
#define fi first
#define se second
#define int long long
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 2e5+10;
int a[N];
int n,v;
bool check(int x)
{
int l = 1,r = x;
while(l < r)
{
//debug2(l,r);
if(a[l] + a[r] > v)return false;
r --;
if(l < r && a[l] + a[r] > v)return false;
l ++;
}
return true;
}
void solve()
{
cin >> n >> v;
int sum = 0;
for(int i = 1;i <= n;i ++){
cin >> a[i];
sum += a[i];
}
sort(a+1,a+n+1);
int l = 1,r = n;
int mid;
while(l < r)
{
mid = l+r+1>> 1;
//debug3(l,r,mid);
if(check(mid))l = mid;
else r = mid-1;
}
cout << sum + n - l + 1<< endl;
}
signed main()
{
/*
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
*/
int T = 1;//cin >> T;
while(T--){
solve();
}
return 0;
}