// Slower C++ solution:
#include <bits/stdc++.h>

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

#define rall(a) a.rbegin(), a.rend()

#define rep(i, s, N) for(ll i{s}; i < N; i++)

#define rem(i, N, s) for(ll i{N}; i > s; i--)

using namespace std;using ll = long long int;using ld = long double;using P = pair<ll,ll>;const int MOD = (int)1e9+7;const string rt = "\n";const string sp = " ";/* 最大公約数 */template<typename T> T gcd(T a, T b){return b != 0 ? gcd(b, a % b) : a;}/* UnionFind */template<typename T> struct UnionFind{vector<T> par;UnionFind(T n) : par(n, -1){}void init(T n){par.assign(n, -1);}T root(T x){if (par[x] < 0) return x;else return par[x] = root(par[x]);}bool issame(T x, T y){return root(x) == root(y);}bool merge(T x, T y){x = root(x); y = root(y);if(x == y) return false;if(par[x] > par[y]) swap(x, y);par[x] += par[y];par[y] = x;return true;}int size(int x) {return -par[root(x)];}};/* コンビネーション */template<typename T> ll combpm(T N_, T C_) {const int NUM_=400001;static ll fact[NUM_+1],factr[NUM_+1],inv[NUM_+1];if (fact[0]==0) {inv[1]=fact[0]=factr[0]=1;for (int i=2;i<=NUM_;++i) inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;for (int i=1;i<=NUM_;++i) fact[i]=fact[i-1]*i%MOD, factr[i]=factr[i-1]*inv[i]%MOD;}if(C_<0 || C_>N_) return 0;return factr[C_]*fact[N_]%MOD*factr[N_-C_]%MOD;}/* 多次元ベクター */template<class T>vector<T> mvec(size_t a){return vector<T>(a);}template<class T, class... Ts>auto mvec(size_t a, Ts... ts){return vector<decltype(mvec<T>(ts...))>(a, mvec<T>(ts...));}/* テスト関数 */void test(ll n){cout << "test" << n << endl;}/* 小数点以下 */void fixsp(ll n){cout << fixed << setprecision(n);}void defsp(ll n){cout << defaultfloat << setprecision(n);}/* 重み付きUnionFind */struct WUnionFind {vector<int> par;vector<int> rank;WUnionFind(int n = 1) {init(n);}void init(int n = 1) {par.resize(n); rank.resize(n);for (int i = 0; i < n; ++i) par[i] = i, rank[i] = 0;}int root(int x) {if (par[x] == x) {return x;}else {int r = root(par[x]);return par[x] = r;}}bool issame(int x, int y) {return root(x) == root(y);}bool merge(int x, int y) {x = root(x); y = root(y);if (x == y) return false;if (rank[x] < rank[y]) swap(x, y);if (rank[x] == rank[y]) ++rank[x];par[y] = x;return true;}};/* IO高速化 */void fastio(){ios::sync_with_stdio(false);cin.tie(NULL);}

/* ここからコード開始 */



int main(){

  fastio;

  string S;

  cin >> S;

  ll N = S.size(); reverse(all(S));

  auto DP = mvec<ll>(N+1,13); DP[0][0] = 1;

  auto d13 = [](ll n) -> ll{

    return (13 + n % 13)%13;

  };

  auto deal = [&](ll i, ll j, ll l) -> void{

    ll k{};

    k = l + d13(j*pow(10, i%3)) * pow(-1, i/3);

    DP[i+1][d13(k)] += DP[i][l]%MOD;

  };

  rep(i, 0, N) rep(l, 0, 13){

    if(S[i] == '?') rep(j, 0, 10) deal(i, j, l);

    else deal (i, S[i] - '0', l);

  }

  cout << DP[N][5]%MOD << rt;

}

// Faster C++ solution for the same problem (measured speedup: 1.158637965579385x):
#include <bits/stdc++.h>

#define rep(i, N) for(int i{};i<N;i++)

using namespace std;

int M = 1e9+7;

int p[100001][13];

string S;

int d(int n){

  return (13 + n%13)%13;

}

void f(int &i, int j, int &l){

  int k{};

  k = l + d(j*(i%3==0?1:i%3==1?10:100)) * (1-i/3%2*2);

  (p[i+1][d(k)] += p[i][l]) %= M;

}

int main(){

  cin >> S;

  reverse(S.begin(), S.end());

  p[0][0] = 1;

  rep(i, S.size()) rep(l, 13){

    if(S[i] == '?') rep(j, 10) f(i, j, l);

    else f(i, S[i] - '0', l);

  }

  cout<<p[S.size()][5]<<"\n";

}