#include #include typedef long long int ll; typedef long long int ull; #define MP make_pair using namespace std; using namespace atcoder; typedef pair P; #define MOD 1000000007 // 1000000007 using mint = modint1000000007; const double pi = 3.1415926536; const int MAX = 2000003; long long fac[MAX], finv[MAX], inv[MAX]; // テーブルを作る前処理 void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } // 二項係数計算 long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll gcd (ll x, ll y) { if (y == 0) return x; else if (y > x) { return gcd (y, x); } else return gcd(x % y, y); } ll lcm (ll x, ll y) { return x / gcd(x, y) * y; } ll my_sqrt(ll x) { ll m = 0; ll M = 1000000001; while (M - m > 1) { ll now = (M + m) / 2; if (now * now <= x) { m = now; } else { M = now; } } return m; } ll keta(ll n) { ll ret = 0; while (n) { n /= 10; ret++; } return ret; } mint dp[10002][3][3][2][2]; int main() { for (int i = 0; i < 10002; i++) { for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) { for (int l = 0; l < 2; l++) { for (int m = 0; m < 2; m++) { dp[i][j][k][l][m] = 0; } } } } } string s; cin >> s; dp[0][0][0][1][0] = 1; for (int i = 0; i < s.length(); i++) { int now = s[i] - '0'; int j2, k2, l2, m2; for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) { for (int l = 0; l < 2; l++) { for (int m = 0; m < 2; m++) { for (int num = 0; num <= 9; num++) { if ((l2 == 1) && (num > now)) continue; if (num < now) { l2 = 0; } else { l2 = l; } if (num == 0) { if (m == 1) continue; j2 = 0; k2 = 0; m2 = 0; } else if (num == 5) { j2 = j; k2 = min(2, k + 1); m2 = 1; } else if ((num == 2) || (num == 6)) { j2 = min(2, j + 1); k2 = k; m2 = 1; } else if ((num == 4) || (num == 8)) { j2 = 2; k2 = k; m2 = 1; } else { j2 = j; k2 = k; m2 = 1; } dp[i + 1][j2][k2][l2][m2] += dp[i][j][k][l][m]; } } } } } } mint ans = dp[s.length()][2][2][0][1] + dp[s.length()][2][2][1][1]; cout << ans.val() << endl; return 0; }