#include<bits/stdc++.h>
#include<fstream>
#define fi(i,a,b) for(int i=(a);i<(b);i++)
#define fd(i,a,b) for(int i=(a);i>(b);--i)
#define llu  long long unsigned
using namespace::std;
 
typedef vector <int> vi;	//The following operations are defined for iterators: get value of an iterator, int x = *it; , increment and decrement iterators it1++, it2--; , compare iterators by '!=' and by '<' , add an immediate to iterator it += 20; <=> shift 20 elements forward , get the distance between iterators, int n = it2-it1; ,  Not only can they operate on any container, they may also perform, for example, range checking and profiling of container usage.
				//The type of iterator can be constructed by a type of container by appending “::iterator”, “::const_iterator”, “::reverse_iterator” or “::const_reverse_iterator” to it. use '!=' instead of '<', and 'empty()' instead of 'size() != 0' -- for some container types, it’s just very inefficient to determine which of the iterators precedes another.
typedef vector<vi> vvi;		//vector< vector<int> > Matrix(N, vector<int>(M, -1));
typedef pair < int, int > ii;	//Pairs are compared first-to-second element. If the first elements are not equal, the result will be based on the comparison of the first elements only; the second elements will be compared only if the first ones are equal. The array (or vector) of pairs can easily be sorted by STL internal functions.(convex hull - For example, if you want to sort the array of integer points so that they form a polygon, it’s a good idea to put them to the vector< pair<double, pair<int,int> >, where each element of vector is { polar angle, { x, y } }. One call to the STL sorting function will give you the desired order of points. )
typedef vector < ii > vii;
typedef vector <vii> vvii;
vector<llu> dp(200020, 0);
int main()
{
  // freopen("in.txt", "r", stdin);
  string s;
vector<llu> arr(200020);
cin>>s;
fi(i, 0, s.size())
    arr[i] = s[i] - '0';
//fi(i, 0, s.size())
//cout<<arr[i];
//cout<<endl;
llu sum = 0;
fi(i, 0, s.size())
{
    dp[i] = arr[i];
    sum = sum % 1000000007;
    sum += dp[i];
 
    fi(j, i+1, s.size())
    {
        dp[j] = ((dp[j-1]% 1000000007)*10 + arr[j] ) % (1000000007);
        sum += dp[j];
//        cout<<i<<' '<<j<<' '<<dp[j]<<endl;
    }
}
cout<<sum<<endl;
return 0;
}
 

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