Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers a1, a2, ..., am is defined as the bitwise XOR of all its elements: CF912B New Year, here CF912B New Year denotes the bitwise XOR operation; more about bitwise XOR can be found ​​​here.​​

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

Input

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).

Output

Output one number — the largest possible xor-sum.

Examples Input Copy

4 3

Output Copy

7

Input Copy

6 6

Output Copy

7

Note

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.

#include
#include
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#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/
ll mul(ll a, ll b) {
    ll ans = 0;
    while (b) {
        if (b & 1)ans = (ans + a);
        b /= 2; a += a;
    }
    return ans;
}


ll qpow(ll a, ll b) {
    ll ans = 1;
    
    while (b) {
        if (b % 2)ans = mul(a, ans);
        b /= 2; a = mul(a, a);
    }
    return ans;
}

ll log2(ll x) {
    int cnt = 0;
    ll tmp = 1;
    while (1) {
        if ((tmp << 1) > x)break;
        tmp <<= 1; cnt++;
    }
    return cnt;
}

int main()
{
    //ios::sync_with_stdio(0);
    ll n, k; rdllt(n); rdllt(k);
    ll ans = 0;
    ll tmp = 1;
    int cnt = 0;
    
    if (k == 1)cout << n << endl;
    else {
        ans = 1;
        while (1) {
            if (ans > n)break;
            ans <<= 1;
        }
        cout << ans - 1 << endl;
    }
    return 0;
}

 

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