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// Arti1990, II UWr

#include <bits/stdc++.h>

#define forr(i, n)                  for(int i=0; i<n; i++)
#define FOREACH(iter, coll)         for(auto iter = coll.begin(); iter != coll.end(); ++iter)
#define FOREACHR(iter, coll)        for(auto iter = coll.rbegin(); iter != coll.rend(); ++iter)
#define lbound(P,K,FUN)             ({auto SSS=P, PPP = P-1, KKK=(K)+1; while(PPP+1!=KKK) {SSS = (PPP+(KKK-PPP)/2); if(FUN(SSS)) KKK = SSS; else PPP = SSS;} PPP;})
#define testy()                     int _tests; scanf("%d", &_tests); FOR(_test, 1, _tests)
#define CLEAR(tab)                  memset(tab, 0, sizeof(tab))
#define CONTAIN(el, coll)           (coll.find(el) != coll.end())
#define FOR(i, a, b)                for(int i=a; i<=b; i++)
#define FORD(i, a, b)               for(int i=a; i>=b; i--)
#define MP                          make_pair
#define PB                          push_back
#define ff                          first
#define ss                          second
#define deb(X)                      X;

#define M 1000000007
#define INF 1000000007LL

using namespace std;

long long n, m;
long long DPO[3007][3007], DPZ[3007][3007];

int solve()
{
    scanf("%lld %lld", &n, &m);
    DPZ[0][0] = 1;
    FOR(i, 1, n)
        FOR(k, 1, i)
        {
            DPO[i][k] = (DPO[i-1][k]*(m-k) + DPZ[i-1][k-1]*(m-k+1)) % M;
            DPZ[i][k] = (DPO[i-1][k] + DPZ[i-1][k])*k % M;
            //cout << i << " " << k << ": DPZ = " << DPZ[i][k] << ", DPO: " << DPO[i][k] << endl;
        }
    long long res = 0;
    FOR(k, 1, n)
        res += DPZ[n][k];
    res %= M;
    printf("%lld\n", res);

    return 0;
}

int main()
{
    //testy()
    solve();

    return 0;
}