#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
typedef long long ll;
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll M, L;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> M >> L;
ll n = 2*M + 1;
vector<ll> a(n), c(n);
for(ll i=0; i<n; ++i) cin >> a[i];
c = a;
ll sum = 0;
for(ll i=0; i<n; ++i) sum += (i-M) * a[i];
// cout << sum << "\n";
//Compute the greedy (c[i] - taken by the greedy)
ll took = 0;
if(sum < L) {
for(ll i=M; i<n; ++i) {
took += (i-M) * a[i];
}
if(took < L) {
cout << "impossible\n";
return 0;
}
for(ll i=M-1; i>=0; --i) {
if(took + (i-M) * a[i] >= L) {
took += (i-M) * a[i];
} else {
c[i] = (took - L)/(M-i);
took += (i-M) * ((took - L)/(M-i));
}
}
} else {
for(ll i=0; i<=M; ++i) {
took += (i-M) * a[i];
}
if(took > L) {
cout << "impossible\n";
return 0;
}
for(ll i=M+1; i<n; ++i) {
if(took + (i-M) * a[i] <= L) {
took += (i-M) * a[i];
} else {
c[i] = (L-took)/(i-M);
took += (i-M) * ((L-took)/(i-M));
}
}
}
// cout << took << "\n";
ll cnt = 0;
for(ll i=0; i<n; ++i) {
cnt += c[i];
// cout << c[i] << " ";
}
// cout << "\n";
//Regret the greedy using dp and fact that we can always maintaing the transition inside tha [L-M, L+M] and obtaining same weight twice can only worsen our solution (for prove look in editorial - https://www.boi2022.de/wp-content/uploads/2022/05/boi2022-day1-vault-spoiler.pdf)
vector<ll> dp(n, -2e18);
ll offset = L-M;
dp[took-offset] = cnt;
auto remove = [&](ll x) {
vector<ll> ndp = dp;
ll L = max(0LL, -x);
ll R = min(n-1, n-1-x);
for(ll i=L; i<=R; ++i) {
ndp[i] = max(dp[i], dp[i+x] - 1);
}
swap(dp, ndp);
};
auto add = [&](ll x) {
vector<ll> ndp = dp;
ll L = max(0LL, x);
ll R = min(n-1, n-1+x);
for(ll i=L; i<=R; ++i) {
ndp[i] = max(dp[i], dp[i-x] + 1);
}
swap(dp, ndp);
};
for(ll i=0; i<n; ++i) {
for(ll j=0; j<min(2*M+5, c[i]); ++j) remove(i-M);
for(ll j=0; j<min(2*M+5, a[i]-c[i]); ++j) add(i-M);
}
if(dp[L - offset]<=0) cout << "impossible\n";
else cout << dp[L-offset] << "\n";
return 0;
}
