Given an integer n, find the maximum value of integer k such that the following condition holds: 

n & (n-1) & (n-2) & (n-3) & ... (k) = 0  where & denotes the [bitwise AND operation.](https://en.wikipedia.org/wiki/Bitwise_operation#AND)

Input

The first line contains a single integer t (1 ≤ t ≤ 3 ⋅ 10^4). Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^9).

Output

For each test case, output a single integer — the required integer k.

Example

Input


3
2
5
17


Output


1
3
15

Note

In the first testcase, the maximum value for which the continuous & operation gives 0 value, is 1.

In the second testcase, the maximum value for which the continuous & operation gives 0 value, is 3. No value greater then 3, say for example 4, will give the & sum 0. 

  * 5   \&   4 ≠ 0, 
  * 5   \&   4   \&   3 = 0. 



Hence, 3 is the answer.
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define mem(a,val) memset(a,(val),sizeof((a)))
#define FAST std::ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define decimal(n)  cout << fixed ; cout << setprecision((n));
#define mp make_pair
#define eb emplace_back
#define f first 
#define s second
#define all(v) v.begin(), v.end()
#define endl "\n"
#define lcm(m,n) (m)*((n)/__gcd((m),(n)))
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(ll (i)=1;(i)<(n);(i)++)
#define repa(i,n,a) for(ll (i)=(a);(i)<(n);(i)++)
#define repr(i,n) for(ll (i)=(n)-1;(i)>=0;(i)--)
#define pll pair<ll,ll>
#define pii pair<int, int>
#define mll map<ll,ll>
#define vll vector<ll>
#define sz(x) (ll)x.size()
#define ub upper_bound
#define lb lower_bound
#define pcnt(x) __builtin_popcountll(x)
const long long N=1e9;
const long long NN=1e18;
const int32_t M=1e9+7;
const int32_t MM=998244353;
template<typename T,typename T1>T maxn(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T minn(T &a,T1 b){if(b<a)a=b;return a;}

void solve()
{
  //code begins from here//
  // ll n=32;
  // //cin>>n;
  // rep1(i,n+1){
  // ll curr=i;
  // ll ii=i;
  // while(curr)
  // {
  //   curr&=(ii-1);
  //   ii--;
  // }
  // cout<<i<<"->"<<ii<<endl;
  // }
  ll n;
  cin>>n;
  repr(i,63)
  {
    if(n&(1LL<<i))
    {
      cout<<(1LL<<i)-1<<endl;
      break;
    }
  }
}

signed main()
{
FAST
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
int testcase=1;
cin>>testcase;
while(testcase--) solve();
return 0;
}
