1 #include <iostream>
2 #include <cstdlib>
3 #include <cstring>
4 #include <queue>
5 #include <cstdio>
6 #include <algorithm>
7 #include <map>
8 #include <time.h>
9 #include <ext/pb_ds/assoc_container.hpp>
10 #include <ext/pb_ds/tree_policy.hpp>
11 #define LL long long
12
13 using namespace std;
14 using namespace __gnu_pbds;
15
16 //U[n]表示3×N棋盘的摆放数，
17 //
18 //V[n]表示缺了1×1边角的3×N棋盘的摆放数
19 //
20 //则
21 //
22 //U[n] = 2*V[n-1] + U[n-2]
23 //
24 //V[n] = U[n-1] + V[n-2]
25 //
26 //其中，
27 //
28 //U[0] = 1, U[2] = 3;
29 //
30 //V[0] = 1, V[1] =1;
31
32 //[U(n) [ 0 2 1 0 [ U(n-1)
33 // V(n) 1 0 0 1 V(n-1)
34 // U(n-1) = 1 0 0 0 * U(n-2)
35 // V(n-1)] 0 1 0 0 ] V(n-2) ]
36 //
37 // [ 0 2 1 0 [ U(1)
38 // = 1 0 0 1 V(1)
39 // 1 0 0 0 ^ (n-1) U(0)
40 // 0 1 0 0 ] V(0) ]
41
42 const int MOD = 12357;
43
44 struct Martix
45 {
46 LL martix[5][5];
47 int row,col;
48 Martix(int _row,int _col)
49 {
50 memset(martix,0,sizeof(martix));
51 row = _row;
52 col = _col;
53 }
54 Martix operator *(const Martix &A)const
55 {
56 Martix C(row, A.col);
57 for(int i = 0; i < row; i++)
58 for(int j = 0; j < A.col; j++)
59 for(int k = 0; k < col; k++)
60 C.martix[i][j] = (C.martix[i][j] + martix[i][k] * A.martix[k][j])%MOD;
61 return C;
62 }
63 };
64
65
66 void solve()
67 {
68 Martix A(4,4), F(4,4), S(4,1);
69 A.martix[0][2] = A.martix[1][0] = A.martix[1][3] = A.martix[2][0] = A.martix[3][1] = 1;
70 F.martix[0][0] = F.martix[1][1] = F.martix[2][2] = F.martix[3][3] = 1;
71 A.martix[0][1] = 2;
72 S.martix[1][0] = S.martix[2][0] = S.martix[3][0] = 1;
73 int n;
74 scanf("%d",&n);
75 if(n&1)
76 {
77 printf("%d\n",0);
78 return ;
79 }
80 n--;
81 while(n > 0)
82 {
83 if(n & 1)
84 F = F*A;
85 A = A*A;
86 n >>= 1;
87 }
88 S = F*S;
89 printf("%lld\n",S.martix[0][0]);
90 }
91
92 int main(void)
93 {
94 solve();
95 return 0;
96 }
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