// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

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

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

#define rep3(i, l, n) for (int i = l; i < n; ++i)

#define chmax(a, b) a = max(a, b)

#define chmin(a, b) a = min(a, b)

#define out(a) cout << a << endl

#define outa(a, n) rep(_, n) cout << a[_] << " "; cout << endl

#define SZ(v) (int)v.size()

#define inf (int)(1e9+7)



ll dp[100001][13];

int main() {

    string s;

    cin >> s;

    fill_n(*dp, 100001 * 13, 0);

    dp[0][0] = 1;

    rep(i, SZ(s)) rep(k, 13) {

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

            rep(l, 10) {

                (dp[i + 1][(k * 10 + l) % 13] += dp[i][k]) %= (ll)(1e9 + 7);

            }

        } else {

            (dp[i + 1][(k * 10 + (s[i] - '0')) % 13] += dp[i][k]) %= (ll)(1e9 + 7);

        }

    }

    out(dp[SZ(s)][5]);

    /*

    rep(i, SZ(s) + 1) {

        rep(j, 13) cout << dp[i][j] << " ";

        cout << endl;

    }

     */

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

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

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

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

const int inf = 1e9 + 7;

const ll INF = 1e18;

int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }

template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }

template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }





void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(10);



    int T;

    // cin >> T;

    T = 1;



    while (T--) {

        solve();

    }

}



int dp[100005][15];

void solve() {

  string s;

  cin >> s;

  int n = sz(s);

  dp[0][0] = 1;

  rep(i, n) rep(j, 13) {

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

      rep(k, 10) {

        add(dp[i + 1][(j * 10 + k) % 13], dp[i][j]);

      }

    } else {

      int k = s[i] - '0';

      add(dp[i + 1][(j * 10 + k) % 13], dp[i][j]);

    }

  }

  cout << dp[n][5] << endl;

}


// Fastest known solution for this problem: