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

const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}


int main(){
  ll N, K; cin >> N >> K;
  ll MOD = 1e9+7;
  ll n = N%MOD;
  ll k = K%MOD;
  mint ans = 0;
  mint A = (mint)(((k+1)*(k+2)-2)%MOD)*mint(2).inv();
  mint B = (mint)((k*(k+1))%MOD)*mint(2).inv();
  ans = A.pow(n) - B.pow(n); 
  cout << ans << endl;
  return 0;
}