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

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
#define int long long

vector<int> a;
int n, k;
vector<int> ans;
int check (int tar) {
int ans = 0;
for (auto t : a) {
if (t <= tar) ans += t;
else ans += tar;
}
return ans;
}
void solve() {
cin >> n >> k;
ll s = 0;
for (int i = 1; i<= n; i ++) {
int t;
cin >> t;
a.pb(t);
s += t;
}
if (s < k) {
puts("-1");
return;
}
int l = 0, r = 1e9;
while (l < r) {
int mid = l + r + 1 >> 1;
if (check(mid) <= k) l = mid;
else r = mid - 1;
}
k -= check(l);
for (int i = 0; i < n; i ++) {
if (a[i] > l) {
ans.pb(i);
}
}
for (int i = k; i < ans.size(); i ++) {
cout << ans[i] + 1 << " ";
}
for (int i = 0; i < min(k, (int)ans.size()); i ++) {
if (a[ans[i]] != l + 1) {
cout << ans[i] + 1 << " ";
}
}
puts("");
}
signed main () {
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}


没有看出这道题是二分。这题我们可以分析单调性发现可以二分。找出最大的轮次然后再对最后一轮进行选择