The Abyss is alive, it evolves itself and generates new errors every 2000 years, which is called the Abyss Cycle.

There're n possible updates of new errors in an Abyss Cycle, and only one of the updates will win out. To find out the winning update, first the Abyss Cycle would mark all the updates. If at one moment only 1 update is marked, then it win out. Otherwise, the Abyss Cycle would perform the procedure below until one update wins out: all the marked updates would randomly choose an integer at range [1,m] with the same possibilities. Let x denotes the minimal chosen number of all the updates, then all the updates with the chosen number equals to x remain marked, and other updates become unmarked.

Please calculate the expected times to perform the procedure. If it expects to perform \dfrac xy times, you should output the value of \dfrac xy\bmod (10^9+7) , where 10^9+7 is a prime.

Input a single line with two integers n,m(1\le n\le10^3,2\le m\le10^3) .

Output a single line with one integer denotes \dfrac xy\bmod(10^9+7) .

In the first example, at the beginning there's only 1 update, so it needn't perform the procedure and the expect times is 0 , and \dfrac01\bmod(10^9+7)=0 .

In the second example, all the possible random-chosen numbers of 2 updates are (1,1),(1,2),(2,1),(2,2) , among which (1,2),(2,1) can find out the one win out, the others cannot and need to perform the procedure again. So that every time, the possibility of perform again is reduce by half, if we calculate the limitation of the geometric sequence, we get: \lim_{n\to\infty}(1+\dfrac12+\dfrac14+\cdots+\dfrac1{2^n})=\lim_{n\to\infty}(2-\dfrac1{2^n})=2 so the expect times is 2 and \dfrac21\bmod(10^9+7)=2 .

In the third example, we can calculate the accurate expect times is \dfrac73 , and \dfrac73\bmod(10^9+7)=333333338 .