// Original C++ solution (slower):
/**

*   @Author : MeGA

*   @InitTime :01/25/20 11:11:36

*   @Idea : 

*

*

*

*

*

*

*

**/

#include <bits/stdc++.h>

using namespace std;

string s;

int n;



const int MOD=1e9+7;

int dp[(int)1e5+7][15];

int solve(int idx,int rem){

    if(idx==n){

       return rem==5?1:0;

    }

    int &ret=dp[idx][rem];

    if(ret!=-1){

    	return  ret;

    }

    ret=0;

    if(s[idx] != '?')

    return ret=solve(idx+1,(rem*10+(s[idx]-'0'))%13);

    for(int i=0;i<=9;i++){

       ret=(1ll*ret+1ll*solve(idx+1,(rem*10+i)%13))%MOD;

    }

    return ret;

}

int main(){

#ifndef ONLINE_JUDGE

  // freopen("in","r",stdin);

  // freopen("out","w",stdout);

#endif

     cin>>s;

     n=s.size();

     memset(dp,-1,sizeof(dp));

     printf("%d",solve(0,0));





}

// Optimized solution to the same problem (faster):
///*****************************************************///

///*.......................Faiyaj Bin Amin............................*///

///*...................................................................*///

///*.........................KUET CSE 2K17......................*///

///*...................................................*///

///*****************************************************///



#include<bits/stdc++.h>

//#include <ext/pb_ds/assoc_container.hpp>

//#include <ext/pb_ds/tree_policy.hpp>



#define fastio      ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)

#define ret return

#define rep(q, i, n) for(int q=i; q<n; q++)

#define forn(i, n) for(int i=0; i<n; i++)

#define mems(a,x) memset(a,x,sizeof(a))

#define ll long long

#define pb push_back

#define em emplace_back

#define mp make_pair

#define all(v) (v).begin(), (v).end()

#define ff first

#define ss second

#define pii pair<int,int>

#define pdd pair<double, double>



#define see(x)        cerr<< __LINE__ << ": " << (#x) << " is "<< (x) <<'\n';

#define seecontent(v) for(auto x : v) { cout << x << " ";} cout << "\n\n";



#define PI 2*acos(0.0)

#define lcm(a, b)  ((a / __gcd(a, b) ) * b)



#define filein freopen ("input.txt", "r", stdin)

#define fileout freopen ("output.txt", "w", stdout)



#define mx 100005

#define INF 1e18+3

#define MOD 1000000007

#define MIN 0



//template <typename T>  using orderedSet =

//    tree<T, null_type, less_equal<T>,

//    rb_tree_tag, tree_order_statistics_node_update>; //find_by_order, order_of_key





/*------------------------------Graph Moves----------------------------*/

//const int fx[]={+1,-1,+0,+0}; // Directional

//const int fy[]={+0,+0,+1,-1};

//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move

//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move

//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move

//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move

/*---------------------------------------------------------------------*/



/**------------------------------------BITMASK------------------------------------------**/

//int Change(int N, int pos){return N= N^(1<<pos);}

//bool Check(int N, int pos){return (bool)(N &(1<<pos));}  /// check bit 0 or 1

/**-----------------------------------------------------------------------------------------------**/



//using namespace __gnu_pbds;

using namespace std;



ll dp[mx][14];

int main()

{

    fastio;

    string s;

    cin >> s;



    dp[0][0]=1;

    forn(i, s.size())

    {

        int fl=100;

        if(s[i]!='?')

            fl=s[i]-'0';

        else

            fl=10;



        forn(j, 10)

        {

            if(fl==10 || fl==j)

            {

                forn(mul, 13)

                {

                    int val = (j+ mul * 10) % 13;

                    dp[i+1][val]+= dp[i][mul];

                    dp[i+1][val] = dp[i+1][val] % MOD;

                }

            }

        }

    }

    cout << dp[s.size()][5] << "\n";



    return 0;

}


// Fastest known solution for this problem: