CF思维联系– CodeForces -CodeForces

Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month).

Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year.

Nastya owns x dresses now, so she became interested in the expected number of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for k + 1 months.

Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109 + 7, because it is easy to see that it is always integer.

Input
The only line contains two integers x and k (0 ≤ x, k ≤ 1018), where x is the initial number of dresses and k + 1 is the number of months in a year in Byteland.

Output
In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109 + 7.

Examples
Input
2 0
Output
4
Input
2 1
Output
7
Input
3 2
Output
21
Note
In the first example a year consists on only one month, so the wardrobe does not eat dresses at all.

In the second example after the first month there are 3 dresses with 50% probability and 4 dresses with 50% probability. Thus, in the end of the year there are 6 dresses with 50% probability and 8 dresses with 50% probability. This way the answer for this test is (6 + 8) / 2 = 7.

这个题画个图就能看出来，如果不考虑最后一天则，前面是个连续的序列。那么最后要求和取平均的过程，换成等差数列求和再取平均。然后化简完就是$(2*a-1)2^{b}+1$，害怕快速幂超时，可以十进制快速幂，也可以欧拉降幂。

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    t f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}

#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 10005
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
const ll phi = 1000000006; //1e9+7的欧拉函数
ll fast_pow(ll a, ll b, ll p)
{
    ll ret = 1;
    for (; b; b >>= 1, a = a * a % p)
        if (b & 1)
            ret = ret * a % p;
    return ret;
}
int main()
{
    ll a, b,c;
    read(a), read(b);
    if (b >= phi)
        b = b % phi + phi; //欧拉降幂
    ll s1 = fast_pow(2, b, MOD);
    cout<<((((2*a)%MOD-1)*s1)+1+MOD)%MOD<<endl;
}