#include<bits/stdc++.h>
using namespace std ;
typedef long long ll ;
const int W = 250005 ;
const int H = 250005 ;
const int WH = 5e5+5 ;
const int P = 1e9+7 ;
int w , h ;
ll c[705][705] ;
vector<vector<ll>> dp ;
vector<ll> dp1 , cnt ;
void init(){
for(int i=0 ; i<705 ; i++){
for(int j=0 ; j<=i ; j++){
if(!i) c[i][j] = 1 ;
else c[i][j] = (c[i-1][j-1]+c[i-1][j])%P ;
}
}
}
ll qpow(ll a , int idx){
if(idx==0) return 1 ;
else if(idx==1) return a ;
if(idx&1) return (a*qpow(a,idx-1))%P ;
ll temp = qpow(a,idx/2) ;
return (temp*temp)%P ;
}
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0) ;
cin >> w >> h ;
if(w==2){
cout << (qpow(2,h)-1+P)%P << '\n' ;
return 0 ;
}
else if(h==2){
if(w==1) cout << "1\n" ;
else if(w==2) cout << "3\n" ;
else cout << "2\n" ;
return 0 ;
}
// this is the solution for time complexity of O(WH^2)
// we need to construct another solution which time complexity is about O(W^2)
// then when WH^2<=W^2 we solve it by using the first solution , else use the second one
if(w*h*h<=w*w || w*h*h<=1e9){ //
init() ;
dp.resize(w+1,vector<ll>(h+1,0LL)) ;
dp[0][0]=1 ;
for(int i=1 ; i<w ; i++){
for(int j=1 ; j<h ; j++){
for(int l=0 ; l<=h-j ; l++) dp[i][j]+=(dp[i-1][l]*c[h-l][j])%P , dp[i][j]%=P ;
}
}
ll ans = 0 ;
for(int j=1 ; j<h ; j++) ans+=dp[w-1][j] ;
cout << ans%P << '\n' ;
}
else{
dp1.resize(w+1) ; cnt.resize(w+1) ; cnt[0] = cnt[1] = 1 ;
for(int i=2 ; i<=w ; i++) cnt[i] = (cnt[i-2]+cnt[i-1])%P ;
for(int i=1 ; i<=w ; i++) cnt[i] = qpow(cnt[i],h) ;
for(int i=1 ; i<=w ; i++){
for(int j=1 ; j<w ; j++){
dp1[i] += (dp1[j]*cnt[w-j])%P ; dp1[i]%=P ;
}
}
cout << dp1[w] << '\n' ;
}
return 0 ;
}
