#include <bits/stdc++.h>
using namespace std;
#define dbgs(x) cerr << (#x) << " --> " << (x) << ' '
#define dbg(x) cerr << (#x) << " --> " << (x) << endl
#define foreach(i,x) for(type(x)i=x.begin();i!=x.end();i++)
#define FOR(ii,aa,bb) for(int ii=aa;ii<=bb;ii++)
#define ROF(ii,aa,bb) for(int ii=aa;ii>=bb;ii--)
#define type(x) __typeof(x.begin())
#define orta (bas + son >> 1)
#define sag (k + k + 1)
#define sol (k + k)
#define pb push_back
#define mp make_pair
#define nd second
#define st first
#define endl '\n'
typedef pair < int ,int > pii;
typedef long long ll;
const long long linf = 1e18+5;
const int mod = (int) 1e9 + 7;
const int logN = 17;
const int inf = 1e9;
const int N = 2e5 + 5;
int cost, n, m, x, y, val[N], root[N], L[N], R[N], a[N], t, k, ans;
int next(int x) { return x == n ? 1 : x + 1; }
int prev(int x) { return x == 1 ? n : x - 1; }
int length(int x, int y) {
if(x < y) return y - x + 1;
return y + n - x + 1;
}
int findset(int x) { return root[x] = root[x] == x ? x : findset(root[x]); }
void merge(int x, int y) {
if((x = findset(x)) == (y = findset(y))) return ;
val[x] = max(val[x], val[y]);
root[y] = x;
R[x] = R[y];
}
int main() {
scanf("%d %d",&n, &k);
FOR(i, 1, n)
scanf("%d",&a[i]);
FOR(i, 1, n) {
root[i] = L[i] = R[i] = i;
val[i] = a[i];
}
priority_queue< pair< int , pii > , vector< pair< int , pii > > , greater< pair< int , pii > > > q;
FOR(i, 1, n) {
if(a[prev(i)] >= a[i] && a[next(i)] >= a[i]) {
q.push(mp(1, mp(i, i)));
}
}
while(!q.empty()) {
x = q.top().nd.st;
y = q.top().nd.nd;
cost = q.top().st;
q.pop();
if(cost > k || length(x, y) == n) continue;
int l = val[findset(prev(x))];
int r = val[findset(next(y))];
int ddd = min(l, r);
if(k < cost * (ll) (ddd - val[findset(x)]))
ddd = val[findset(x)] + (k / cost);
ans += (ddd - val[findset(x)]) * 2;
k -= (ddd - val[findset(x)]) * cost;
val[findset(x)] = ddd;
if(l == ddd) merge(prev(x), x);
if(r == ddd) merge(x, next(y));
x = L[findset(x)], y = R[findset(x)];
if(val[findset(prev(x))] >= val[findset(x)] && val[findset(next(y))] >= val[findset(x)] && length(x, y) > cost)
q.push(mp(length(x, y), mp(x, y)));
}
cout << ans << endl;
return 0;
}
