Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

Source

 1
 
 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int N=30;
 5 int n,k,mod;
 6 struct node{
 7     int a[N][N];
 8     void clr(){memset(a,0,sizeof(a));}
 9     void init(){for (int i=0;i<n;++i) a[i][i]=1;}
10 }base,ans;
11 node operator + (node a,node b)
12 {
13     node c; c.clr();
14     for (int i=0;i<n;++i)
15     for (int j=0;j<n;++j)
16     c.a[i][j]=(a.a[i][j]+b.a[i][j])%mod;
17     return c;
18 }
19 node operator * (node a,node b)
20 {
21     node c; c.clr();
22     for (int i=0;i<n;++i)
23     for (int j=0;j<n;++j)
24     for (int k=0;k<n;++k)
25     c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mod;
26     return c;
27 }
28 node Pow(node now,int x)
29 {
30     if (x==1) return now;
31     node tmp; --x;
32     for (tmp=now;x;x>>=1,now=now*now)
33     if (x&1) tmp=tmp*now;
34     return tmp;
35 }
36 node S(node now,int x)
37 {
38     if (x==1) return now;
39     node tmp; tmp.clr(); tmp.init();
40     tmp=tmp+Pow(now,x/2);
41     tmp=tmp*S(now,x/2);
42     if (x%2) tmp=tmp+Pow(now,x);
43     return tmp;
44 }
45 int main()
46 {
47     scanf("%d%d%d",&n,&k,&mod);
48     base.clr();
49     for (int i=0;i<n;++i)
50     for (int j=0;j<n;++j)
51     scanf("%d",&base.a[i][j]);
52     ans=S(base,k);
53     for (int i=0;i<n;++i)
54     {
55         for (int j=0;j<n;++j)
56         printf("%d ",ans.a[i][j]);
57         printf("\n");
58     }
59     return 0;
60 }

POJ 3233 Matrix Power Series