#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <iostream>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const LL mod = 1e7 + 7;const int maxn = 15;struct Matrix {    int n, m;    LL data[maxn][maxn];    Matrix(int n = 0, int m = 0): n(n), m(m) {        memset(data, 0, sizeof(data));    }    void print() {        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= m; j++) {                cout << data[i][j] << " ";            }            cout << endl;        }    }};Matrix operator * (Matrix a, Matrix b) {    Matrix ret(a.n, b.m);    for(int i = 1; i <= a.n; i++) {        for(int j = 1; j <= b.m; j++) {            for(int k = 1; k <= a.m; k++) {                ret.data[i][j] += a.data[i][k] * b.data[k][j];                ret.data[i][j] %= mod;            }        }    }    return ret;}Matrix pow(Matrix mat, LL p) {    Matrix ret(mat.n, mat.m);    if(p == 0) {        for(int i = 1; i <= mat.n; i++) ret.data[i][i] = 1;        return ret;    }    if(p == 1) return mat;    ret = pow(mat * mat, p / 2);    if(p & 1) ret = ret * mat;    return ret;}int main() {    Matrix A(12, 12); A.data[1][1] = 1;    for(int i = 2; i <= 12; i++) {        A.data[i][1] = 1; A.data[i][2] = 10;    }    for(int i = 3; i <= 12; i++) {        for(int j = 1, k = 3; j < i - 1; j++, k++) {            A.data[i][k] = 1;        }    }    A.print();    LL n, m;    while(cin >> n >> m) {        LL a[11] = {0};        for(int i = 1; i <= n; i++) cin >> a[i];        Matrix A0(12, 1);        A0.data[1][1] = 3; A0.data[2][1] = 233;        for(int i = 3, j = 1; i <= 12; i++, j++) {            A0.data[i][1] = a[j] + A0.data[i - 1][1];        }        Matrix ans = pow(A, m - 1) * A0;        cout << ans.data[n + 2][1] << endl;    }    return 0;}