#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define ll long long #define fi first #define se second #define lengthof(x) (sizeof(x) / sizeof(*(x))) //fill((int*)d, (int*)(d + lengthof(d)), -1); using namespace std; string a,b; const ll MOD=(ll)1e9+7; ll dp[10010][2][2][3][8]; inline ll solve(string s){ int n=(int)s.length(); dp[0][0][0][0][0]=1; for(int i=0;i=0;i--){ if(s[i]-'0'-1<0){ s[i]='9'; }else{ s[i]-=1; break; } } for(int i=0;i<(int)s.length();i++){ if(s=="0")break; if(s[i]=='0')s=s.substr(1,s.length()-1); else break; } return s; } int main(){ ios::sync_with_stdio(false); cin.tie(0); cout.precision(10); cout<>a>>b; ll ans=solve(b); for(int i=0;i<(int)a.length()+5;i++){ for(int j=0;j<2;j++)for(int k=0;k<2;k++)for(int l=0;l<3;l++)for(int m=0;m<8;m++)dp[i][j][k][l][m]=0; } //cout<