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>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,_7,_8,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
#define prArr(a) {cerr<<(#a)<<"={";int i=0;for(auto t:(a))cerr<<(i++?", ":"")<<t;cerr<<"}"<<endl;}
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<60)+1e9; // ~ 1.15 * 1e18
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){
  return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
//INSERT ABOVE HERE

string S, T;
vector<Int> B;

Int dfs2(Int idx, Int same){
  static Int mem[200010][2], used[200010][2];
  if(idx == (Int)T.size()) return 1;
  if(used[idx][same]++) return mem[idx][same];

  Int res = 0;
  for(Int num=0;num<10;num++){
    if(isdigit(T[idx]) && num != T[idx] - '0') continue;
    if(same && S[idx] - '0' < num) continue;
    Int nsame = same & (S[idx] - '0' == num);
    res += dfs2(idx + 1, nsame);
    res %= mod;
  }
  return mem[idx][same] = res;
}

Int dfs(Int idx, Int same, Int keta){
  static Int mem[200010][2], used[200010][2];
  if(idx == (Int)T.size()) return 0;
  if(used[idx][same]++) return mem[idx][same];
  Int res = 0, update = 0;
  for(Int num=0;num<10;num++){
    if(isdigit(T[idx]) && num != T[idx] - '0') continue;
    if(same && S[idx] - '0' < num) continue;
    Int nsame = same && (S[idx] - '0' == num);
    Int t = dfs(idx + 1, nsame, keta-1);
    if(t == -1) continue;
    update = 1;
    res += B[keta] * num * dfs2(idx + 1, nsame) + t;
    res %= mod;
  }
  if(!update) res = -1;
  return mem[idx][same] = res;
}


signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  cin>>S;
  cin>>T;
  {
    reverse(S.begin(), S.end());
    reverse(T.begin(), T.end());
    while(S.size() < T.size()) S += '0';
    while(T.size() < S.size()) T += '0';
    reverse(S.begin(), S.end());
    reverse(T.begin(), T.end());
  }

  const Int N = 200010;
  B.resize(N);
  B[0] = 1;
  for(Int i=1;i<N;i++) B[i] = B[i-1] * 10 % mod;
  Int ans = dfs(0, 1, T.size() - 1);
  if(ans == -1) ans = 0;
  cout<<ans<<endl;


  return 0;
}

