D: Many Decimal Integers

problem

Given a string S consisting only of numbers (0-9) and a string T consisting only of numbers and `?`. S and T are the same length.

Consider changing each `?` That exists in T to one of the numbers from 0 to 9 to create the string T'consisting of only numbers. At this time, it must be f (T') \ leq f (S). Where f (t) is a function that returns an integer value when the string t is read as a decimal number. Also, the number in the most significant digit of T'may be 0.

For all possible strings T', find the remainder of the sum of the values ​​of f (T') divided by 10 ^ 9 + 7. If there is no T'that meets the conditions, answer 0.

Input format


S
T


Constraint

* 1 \ leq | S | = | T | \ leq 2 \ times 10 ^ 5
* S is a string consisting only of numbers (0 to 9)
* T is a string consisting only of numbers and `?`



Output format

Divide the sum of T's that satisfy the condition by 10 ^ 9 + 7 and output the remainder on one line.

Input example 1


73
6?


Output example 1


645

There are 10 possible strings for T', from 60 to 69. The sum of these is 645.

Input example 2


42
? 1


Output example 2


105

The number in the most significant digit of T'can be 0, so 01 also satisfies the condition.

Input example 3


1730597319
16 ?? 35 ?? 8?


Output example 3


502295105

Find the remainder divided by 10 ^ 9 + 7.





Example

Input

73
6?


Output

645
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 2000000000
#define LLINF 1000000000000000ll
#define sz(x) ((int)(x).size())
#define fi first
#define sec second
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define repn(i,a,n) for(int (i)=(a);(i)<(int)(n);(i)++)
#define EQ(a,b) (abs((a)-(b))<eps)
#define dmp(x) cerr << __LINE__ << " " << #x << " " << x << endl;
template<class T> void chmin(T& a,const T& b){if(a>b)a=b;}
template<class T> void chmax(T& a,const T& b){if(a<b)a=b;}
template<class T,class U>
ostream& operator << (ostream& os,pair<T,U>& p){
	os << p.fi << ',' << p.sec; return os;
}
template<class T,class U>
istream& operator >> (istream& is,pair<T,U>& p){
	is >> p.fi >> p.sec; return is;
}
template<class T>
ostream& operator << (ostream &os,const vector<T> &vec){
	for(int i=0;i<vec.size();i++){
		os << vec[i];
		if(i+1<vec.size())os << ' ';
	}
	return os;
}
template<class T>
istream& operator >> (istream &is,vector<T>& vec){
	for(int i=0;i<vec.size();i++)is >> vec[i];
	return is;
}
void fastio(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
}
// How to devide :
// ModInt a(6ll);
// ModInt b(2ll);
// a *= b.exp(MOD-2ll);  -> a/=b;  result: a = 3
ll MOD = 1000000007ll; // if inv is needed, this shold be prime.
struct ModInt{
	ll val;
	ModInt():val(0ll){}
	ModInt(ll v):val(((v%MOD)+MOD)%MOD){}
	ModInt exp(ll y)const{
		if(!y)return ModInt(1ll);
		ModInt a = exp(y/2ll);
		a *= a;
		if(y&1)a*=(*this);
		return a;
	}
	bool operator==(const ModInt& x)const{return val==x.val;}
	inline bool operator!=(const ModInt& x)const{return !(*this==x);}
	bool operator<(const ModInt& x)const{return val<x.val;}
	bool operator>(const ModInt& x)const{return val>x.val;}
	inline bool operator>=(const ModInt& x)const{return !(*this<x);}
	inline bool operator<=(const ModInt& x)const{return !(*this>x);}
	ModInt& operator+=(const ModInt& x){if((val+=x.val)>=MOD)val-=MOD;return *this;}
	ModInt& operator-=(const ModInt& x){if((val+=MOD-x.val)>=MOD)val-=MOD;return *this;}
	ModInt& operator*=(const ModInt& x){(val*=x.val)%=MOD;return *this;}
	ModInt operator+(const ModInt& x)const{return ModInt(*this)+=x;}
	ModInt operator-(const ModInt& x)const{return ModInt(*this)-=x;}
	ModInt operator*(const ModInt& x)const{return ModInt(*this)*=x;}
};
istream& operator>>(istream&i,ModInt&x){i>>x.val;return i;}
ostream& operator<<(ostream&o,const ModInt&x){o<<x.val;return o;}
ModInt pow(ModInt a,ll x){
	ModInt res = ModInt(1ll);
	while(x){
		if(x&1)res *= a;
		x >>= 1;
		a = a*a;
	}
	return res;
}
const int SIZE = 100100;
ModInt inv[SIZE+10],fac[SIZE+10],facinv[SIZE+10];
// notice: 0C0 = 1 
ModInt nCr(int n,int r){
	assert(!(n<r));
	assert(!(n<0||r<0));
	return fac[n]*facinv[r]*facinv[n-r];
}
void init(){
	fac[0]=ModInt(1ll);
	for(int i=1;i<=SIZE;i++)fac[i]=fac[i-1]*ModInt(i);
	inv[1]=ModInt(1ll);
	for(int i=2;i<=SIZE;i++)inv[i]=ModInt(0ll)-ModInt(MOD/i)*inv[MOD%i];
	facinv[0]=ModInt(1ll);
	for(int i=1;i<=SIZE;i++)facinv[i]=facinv[i-1]*inv[i];
	return;
}
string s,t;
ModInt dp[200100][2];
ModInt ep[200100][2];
int main(){
  cin >> s;
  cin >> t;
  ep[0][1]=ModInt(1ll);
  for(int i=0;i<t.size();i++){
    if(t[i]=='?'){
      dp[i+1][0] += dp[i][0]*ModInt(100ll)+ep[i][0]*ModInt(45ll);
      ep[i+1][0] += ep[i][0]*ModInt(10ll);
      int c = (int)(s[i]-'0');
      int sum = 0;
      for(int j=0;j<c;j++)sum += j;
      dp[i+1][0] += dp[i][1]*ModInt(10*c)+ep[i][1]*ModInt(sum);
      ep[i+1][0] += ep[i][1]*ModInt(c);
      dp[i+1][1] += dp[i][1]*ModInt(10ll)+ep[i][1]*ModInt(c);
      ep[i+1][1] += ep[i][1];
    }else{
      int sc = (int)(s[i]-'0');
      int tc = (int)(t[i]-'0');
      if(tc==sc){
        dp[i+1][1] += dp[i][1]*ModInt(10ll)+ep[i][1]*ModInt(tc);
        ep[i+1][1] += ep[i][1];
      }else if(tc<sc){
        dp[i+1][0] += dp[i][1]*ModInt(10ll)+ep[i][1]*ModInt(tc);
        ep[i+1][0] += ep[i][1];
      }
      dp[i+1][0] += dp[i][0]*ModInt(10ll)+ep[i][0]*ModInt(tc);
      ep[i+1][0] += ep[i][0];
    }
    // cout << dp[i+1][0] << ' ' << dp[i+1][1] << endl; 
  }
  cout << dp[t.size()][0]+dp[t.size()][1] << endl;
	return 0;
}

