Problem:
Score : 1000 points

Problem Statement

Takahashi and Snuke came up with a game that uses a number sequence, as follows:

Prepare a sequence of length M consisting of integers between 0 and 2^N-1 (inclusive): a = a_1, a_2, \ldots, a_M.

Snuke first does the operation below as many times as he likes:

Choose a positive integer d, and for each i (1 \leq i \leq M), in binary, set the d-th least significant bit of a_i to 0. (Here the least significant bit is considered the 1-st least significant bit.)

After Snuke finishes doing operations, Takahashi tries to sort a in ascending order by doing the operation below some number of times. Here a is said to be in ascending order when a_i \leq a_{i + 1} for all i (1 \leq i \leq M - 1).

Choose two adjacent elements of a: a_i and a_{i + 1}. If, in binary, these numbers differ in exactly one bit, swap a_i and a_{i + 1}.

There are 2^{NM} different sequences of length M consisting of integers between 0 and 2^N-1 that can be used in the game.

How many among them have the following property: if used in the game, there is always a way for Takahashi to sort the sequence in ascending order regardless of Snuke's operations? Find the count modulo (10^9 + 7).

Constraints

All values in input are integers.

1 \leq N \leq 5000

2 \leq M \leq 5000

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number, modulo (10^9 + 7), of sequences with the property: if used in the game, there is always a way for Takahashi to sort the sequence in ascending order regardless of Snuke's operations.

Sample Input 1

2 5

Sample Output 1

352

Consider the case a = 1, 3, 1, 3, 1 for example.

When the least significant bit of each element of a is set to 0, a = 0, 2, 0, 2, 0;

When the second least significant bit of each element of a is set to 0, a = 1, 1, 1, 1, 1;

When the least two significant bits of each element of a are set to 0, a = 0, 0, 0, 0, 0.

In all of the cases above and the case when Snuke does no operation to change a, we can sort the sequence by repeatedly swapping adjacent elements that differ in exactly one bit.
Thus, this sequence has the property: if used in the game, there is always a way for Takahashi to sort the sequence in ascending order regardless of Snuke's operations.

Sample Input 2

2020 530

Sample Output 2

823277409

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=5000+100;
const db pi=acos(-1.0);
#define lowbit(x) ((x)&(-x))
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 1000000007
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

namespace My_Math{
    #define N 100000

    int fac[N+100],invfac[N+100];

    int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
    int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
    int mul(int x,int y) {return 1ll*x*y%maxd;}
    ll qpow(ll x,int y)
    {
        ll ans=1;
        while (y)
        {
            if (y&1) ans=mul(ans,x);
            x=mul(x,x);y>>=1;
        }
        return ans;
    }
    int getinv(int x) {return qpow(x,maxd-2);}

    int C(int n,int m)
    {
        if ((n<m) || (n<0) || (m<0)) return 0;
        return mul(mul(fac[n],invfac[m]),invfac[n-m]);
    }

    void math_init()
    {
        fac[0]=invfac[0]=1;
        rep(i,1,N) fac[i]=mul(fac[i-1],i);
        invfac[N]=getinv(fac[N]);
        per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
    }
    #undef N
}
using namespace My_Math;

int n,m,f[N][N];

int main()
{
    n=read();m=read();
    rep(i,1,m) f[0][i]=1;
    rep(i,1,n)
    {
        int sum=0;
        rep(j,1,m)
        {
            f[i][j]=add(sum,mul(j+1,f[i-1][j]));
            sum=add(mul(sum,2),mul(f[i-1][j],j));
        }
    }
    printf("%d",f[n][m]);
    return 0;
}