#include <stdio.h>
#include <cstdlib>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <string>
#include <memory.h>
#include <map>
#define _USE_MATH_DEFINES
#include <math.h>
#include <list>
#include <fstream>
#include <time.h>
#include <cassert>

using namespace std;
 
#define FOR(i,a,b) for (int i = a; i <= b; i++)
#define REP(i,n) FOR(i,0,n-1)
#define CLR(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define mp make_pair
#define Inf 1000000000
#define Mi 1000000007
#define N 70
 
typedef double ld;
typedef unsigned long long ll;
typedef int i;
typedef vector<i> vi;
typedef vector<vi> vvi;

int min (int a, int b) {if (a < b) return a; return b;}
int Abs(int a) {if (a < 0) return -a; else return a;}

//class for matrix
class TMatr{
public:
    int n;//length and width of matrix. cause the matrix is quadratic it's the same
    ll A[N][N];
    TMatr operator * (TMatr b);
};

TMatr res;

TMatr TMatr::operator * (TMatr b)//operator of multiplying matrixes
{
    res.n = n;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++){
            res.A[i][j] = 0;
            for (int k = 0; k < n; k++)
                res.A[i][j] = (res.A[i][j] + (this->A[i][k] * b.A[k][j]) % Mi) % Mi; //res[i][j] = sum(a[i][k]*b[k][j]), k = 0..n-1 
        }
    return res;
}

bool bit[4];
//this function returns true if we can put two masks together and false otherway
bool CanGo (int mask1, int mask2, int len)//mask1 and mask2 - two masks, len - their length
{
    for (int i = 0; i < len - 1; i++)
    {
        //the ith bit from the end of first mask(numeration from 0)
        bit[0] = (mask1 >> i) & 1;
        //the i+1th bit from the end of first mask(numeration from 0)
        bit[1] = (mask1 >> (i + 1)) & 1;
        //the ith bit from the end of second mask(numeration from 0)
        bit[2] = (mask2 >> i) & 1;
        //the i+1th bit from the end of second mask(numeration from 0)
        bit[3] = (mask2 >> (i + 1)) & 1;
        if (bit[0] && bit[1] && bit[2] && bit[3]) return false; //if all bits are 1 (cells are black) return false
        if (!bit[0] && !bit[1] && !bit[2] && !bit[3]) return false;//if all bits are 0 (cells are white) return false too
    }
    return true;//if there weren't rectangle 2*2 of all bits 0 or all bits 1 we should return true
}

TMatr A;
TMatr B;
int n;
ll m;

int main ()
{
    //freopen ("input.txt", "r", stdin);
    //freopen ("output.txt", "w", stdout);
    cin >> n >> m;
    //assert (n >= 1 && n <= 6);
    //assert (m >= 1 && m < (1LL << 63));
    int k = 1 << n;
    A.n = k;
    for (int m1 = 0; m1 < k; m1++)
        for (int m2 = 0; m2 < k; m2++)
            A.A[m1][m2] = CanGo(m1, m2, n);//fill the matrix D. D[i][j] = true if we can put masks i and j together and false otherway
    B.n = k;
    CLR(B.A, 0);
    for (int i = 0; i < k; i++)
        B.A[i][i] = 1;//single matrix
    m--;
    //binary power
    while (m){
        if (m & 1)
            B = B * A;
        A = A * A;
        m >>= 1;
    }
    ll ans = 0;
    //answer is sum of values in all cells of result matrix
    for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
            ans = (ans + B.A[i][j]) % Mi;
    printf ("%d\n", ans);
    return 0;
}