1 #include <bits/stdc++.h>
2
3 using namespace std;
4 #define rep(i,a,n) for (long long i=a;i<n;i++)
5 #define per(i,a,n) for (long long i=n-1;i>=a;i--)
6 #define pb push_back
7 #define mp make_pair
8 #define all(x) (x).begin(),(x).end()
9 #define fi first
10 #define se second
11 #define SZ(x) ((long long)(x).size())
12 typedef vector<long long> VI;
13 typedef long long ll;
14 typedef pair<long long,long long> PII;
15 const ll mod=1e9+7;
16 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
17 // head
18
19 long long _,n;
20 namespace linear_seq
21 {
22 const long long N=10010;
23 ll res[N],base[N],_c[N],_md[N];
24
25 vector<long long> Md;
26 void mul(ll *a,ll *b,long long k)
27 {
28 rep(i,0,k+k) _c[i]=0;
29 rep(i,0,k) if (a[i]) rep(j,0,k)
30 _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
31 for (long long i=k+k-1;i>=k;i--) if (_c[i])
32 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
33 rep(i,0,k) a[i]=_c[i];
34 }
35 long long solve(ll n,VI a,VI b)
36 { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
37 // printf("%d\n",SZ(b));
38 ll ans=0,pnt=0;
39 long long k=SZ(a);
40 assert(SZ(a)==SZ(b));
41 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
42 Md.clear();
43 rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
44 rep(i,0,k) res[i]=base[i]=0;
45 res[0]=1;
46 while ((1ll<<pnt)<=n) pnt++;
47 for (long long p=pnt;p>=0;p--)
48 {
49 mul(res,res,k);
50 if ((n>>p)&1)
51 {
52 for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
53 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
54 }
55 }
56 rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
57 if (ans<0) ans+=mod;
58 return ans;
59 }
60 VI BM(VI s)
61 {
62 VI C(1,1),B(1,1);
63 long long L=0,m=1,b=1;
64 rep(n,0,SZ(s))
65 {
66 ll d=0;
67 rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
68 if (d==0) ++m;
69 else if (2*L<=n)
70 {
71 VI T=C;
72 ll c=mod-d*powmod(b,mod-2)%mod;
73 while (SZ(C)<SZ(B)+m) C.pb(0);
74 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
75 L=n+1-L; B=T; b=d; m=1;
76 }
77 else
78 {
79 ll c=mod-d*powmod(b,mod-2)%mod;
80 while (SZ(C)<SZ(B)+m) C.pb(0);
81 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
82 ++m;
83 }
84 }
85 return C;
86 }
87 long long gao(VI a,ll n)
88 {
89 VI c=BM(a);
90 c.erase(c.begin());
91 rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
92 return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
93 }
94 };
95
96 int main()
97 {
98 while(~scanf("%I64d", &n))
99 { printf("%I64d\n",linear_seq::gao(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205},n-1));
100 }
101 }
