In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a _0,1 = 233,a _0,2 = 2333,a _0,3 = 23333...) Besides, in 233 matrix, we got a _i,j = a _i-1,j +a _i,j-1( i,j ≠ 0). Now you have known a _1,0,a_2,0,...,a _n,0, could you tell me a _n,m in the 233 matrix?

1 1
1
2 2
0 0
3 7
23 47 16

234
2799
72937

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 18
#define N 33
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-9;
const double PI = acos(-1.0);
/*
组合数学 找规律
递归显然不行，列数太多
只需考虑每个点被加上的次数
a(i,0) = a(i,1) 到 a(n,m) 路径条数（向左和向下两个方向） C(n+m-i-1,n)    
发现列数太多没办法打表 再换一种方法
矩阵快速幂
从第一列向后考虑 找出他们的转移矩阵（这里很巧妙的加了一条边 凑2333后面的3）十分巧妙！~
*/
LL a[MAXN], n, m;
struct mat
{
    LL data[MAXN][MAXN];
    mat()
    {
        memset(data, 0, sizeof(data));
    }
    mat operator*(const mat& rhs)
    {
        mat ret;
        for (int i = 1; i <= n + 2; i++)
        {
            for (int j = 1; j <= n + 2; j++)
            {
                for (int k = 1; k <= n + 2; k++)
                    ret.data[i][j] = (ret.data[i][j] + data[i][k] * rhs.data[k][j]) % MOD;
            }
        }
        return ret;
    }
};
mat fpow(mat a, LL b)
{
    if (b <= 0) return a;
    mat tmp = a, ret;
    for (int i = 1; i <= n + 2; i++)
        ret.data[i][i] = 1;
    while (b!= 0)
    {
        if (b & 1)
            ret = tmp*ret;
        tmp = tmp*tmp;
        b = b / 2;
    }
    return ret;
}
int main()
{
    while (cin >> n >> m)
    {
        a[1] = 23;
        for (int i = 2; i <= n + 1; i++)
            cin >> a[i];
        a[n + 2] = 3;
        mat ans;
        for (int i = 1; i <= n + 1; i++)
        {
            ans.data[i][1] = 10;
            ans.data[i][n + 2] = 1;
            for (int j = 2; j <= i; j++)
                ans.data[i][j] = 1;
        }
        ans.data[n + 2][n + 2] = 1;
        ans = fpow(ans, m);
        LL result = 0;
        for (int i = 1; i <= n + 2; i++)
            result = (result + a[i] * ans.data[n + 1][i]) % MOD;
        cout << result << endl;
    }
    return 0;
}