Finales are born to be exciting. Performers play hard to draw audiences’ attention and then take a perfect curtain call. As the last problem and the finale of the problem set, however, we want you to recall a simple algorithm. Like me, it may be the first algorithm you’ve learned, called Bubble Sort.

void bubble_sort ( int a [] , int n ) { // 0 - based , sort from lowest to highest
	for ( int i = 1; i < n ; i ++) {
		for ( int j = 0; j < n - i ; j ++) {
			if ( a [ j ] > a [ j + 1]) {
				swap ( a [ j ] , a [ j + 1]);
			}
		} // after i - th inner iteration , a [ n - i ] is correct
	}
}

Given a permutation of length n, as you might know, Bubble Sort runs in \Omega(n^2) in the worst case. It’s quite a traditional idea to count the number of calls of “swap” in the algorithm. As you are stronger now, you want to count that number in a dynamic permutation with the following events that might happen:

• Reverse the permutation, meaning that the permutation is replaced with

• Shift the permutation to the left by 1, meaning that the permutation is replaced with

All you need to do is to output the number of “swap” that would be called if we sort the permutation with the above Bubble Sort code after each operation.

The first line contains two integers n , m (1 \leq n \leq 3 \times 10^5,1 \leq m \leq 6 \times 10^5), denoting the length of permutation and the number of operations.

The second line contains n integers separated by spaces, and the i-th denotes the initial p_i.

The third line contains a single string containing m letters consisting of R and S. The i-th letter denotes the i-th operation, where R or S denotes the Reverse or Shift operation, respectively.

It's guaranteed that p forms a correct permutation of 1,2,\dots,n.

In the first line, print the number of "swap" would be called when Bubble Sort the initial p.

In the second line, print a single string of m digits. The i-th denotes the number of "swap" would be called to Bubble Sort the permutation, modulo 10.