Problem: 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.

Examples

Input

2 5


Output

352


Input

2020 530


Output

823277409

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 
int fstTrue(function<bool(int)> f, int lo, int hi) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		int mid = (lo+hi)/2; 
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
template<class A> str ts(complex<A> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) { 
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
	bool fst = 1; str res = "{";
	for (const auto& x: v) {
		if (!fst) res += ", ";
		fst = 0; res += ts(x);
	}
	res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
	return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << ts(h); if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

struct mi {
	typedef decay<decltype(MOD)>::type T; 
 	/// don't silently convert to T
	T v; explicit operator T() const { return v; }
	mi() { v = 0; }
	mi(ll _v) { 
		v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.v < b.v; }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }
	friend str ts(mi a) { return ts(a.v); }
   
	mi& operator+=(const mi& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mi& operator*=(const mi& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mi operator-() const { return mi(-v); }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

vector<vmi> scmb; // small combinations
void genComb(int SZ) {
	scmb.assign(SZ,vmi(SZ)); scmb[0][0] = 1;
	FOR(i,1,SZ) F0R(j,i+1) 
		scmb[i][j] = scmb[i-1][j]+(j?scmb[i-1][j-1]:0);
}

int N,M;
mi dp[5001][5001];
mi po2[5001];

int main() {
	po2[0] = 1; FOR(i,1,5001) po2[i] = 2*po2[i-1];
	setIO(); re(N,M);
	FOR(j,1,M+1) dp[0][j] = 1;
	FOR(i,1,N+1) {
		mi cum = 0;
		FOR(j,1,M+1) {
			cum *= 2;
			cum += (j-1)*dp[i-1][j-1];
			dp[i][j] += (j+1)*dp[i-1][j];
			dp[i][j] += cum;
			// for (int len = 2;len <= j; ++len)
			// 	dp[i][j] += po2[len-2]*(j-len+1)*dp[i-1][j-len+1];
		}
	}
	ps(dp[N][M]);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/