Problem: Valera has a strip infinite in both directions and consisting of cells. The cells are numbered by integers. The cell number 0 has a robot.

The robot has instructions — the sequence of moves that he must perform. In one move, the robot moves one cell to the left or one cell to the right, according to instructions. Before the robot starts moving, Valera puts obstacles in some cells of the strip, excluding cell number 0. If the robot should go into the cell with an obstacle according the instructions, it will skip this move.

Also Valera indicates the finish cell in which the robot has to be after completing the entire instructions. The finishing cell should be different from the starting one. It is believed that the robot completed the instructions successfully, if during the process of moving he visited the finish cell exactly once — at its last move. Moreover, the latter move cannot be skipped.

Let's assume that k is the minimum number of obstacles that Valera must put to make the robot able to complete the entire sequence of instructions successfully and end up in some finishing cell. You need to calculate in how many ways Valera can choose k obstacles and the finishing cell so that the robot is able to complete the instructions successfully.

Input

The first line contains a sequence of characters without spaces s1s2... sn (1 ≤ n ≤ 106), consisting only of letters "L" and "R". If character si equals "L", then the robot on the i-th move must try to move one cell to the left. If the si-th character equals "R", then the robot on the i-th move must try to move one cell to the right.

Output

Print a single integer — the required number of ways. It's guaranteed that this number fits into 64-bit signed integer type.

Examples

Input

RR


Output

1


Input

RRL


Output

1

Note

In the first sample Valera mustn't add any obstacles and his finishing cell must be cell 2.

In the second sample, Valera must add an obstacle in cell number 1, and his finishing cell must be cell number  - 1. In this case robot skips the first two moves and on the third move he goes straight from the starting cell to the finishing one. But if Valera doesn't add any obstacles, or adds an obstacle to another cell, then the robot visits the finishing cell more than once.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main
{
    static char[] inp = null;
    static int len;
    public static void main(String[] args)
    {
        // get input
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
            inp = reader.readLine().toCharArray();
        } catch (Exception e) {System.out.println(e.toString());};

        // swap L, R if the last element is L
        len = inp.length;
        if (inp[len - 1] == 'L')
            for (int i = 0; i < len; i++)
                inp[i] = inp[i] == 'R' ? 'L' : 'R';

//        System.out.println(Arrays.toString(inp));
        // check if this simulation works 
        // (place an obstacle at -1e7
        // if it works, then print 0 and stop
        if (simulate((int) -1e7))
        {
            System.out.println(1);
            return;
        }

        // otherwise do binary search
        // the target is 2 numbers
        // tartget_low, tartget_high.
        // s.t. target_low works, but target_low - 1 does not work
        // s.t. target_high works, but target_high + 1 does not work
        // it must be b/w - 1e6 to 0
        int lb = lower_bound(-1, -5 + (int) -1e6);
        int hb = higher_bound(0, -5 + (int) -1e6);

        // then print target_high - target_low + 1
//        System.out.println(hb - lb + 1);
//        System.out.println(hb);
        if (simulate(lb))
            System.out.println(-1 * lb);
        else
            System.out.println(0);
    }
    // lower_bound high low
    static int lower_bound(int high, int low)
    {
//        System.out.println(high + " : " + low);
        if (high - low == 1)
        {
            return high;
        }
        // middle = (high + low + 1) / 2
        int middle = (high + low) / 2;
        if (simulate(middle))
        {
            // if middle works, then go to lower_bound(middle, low)
            return lower_bound(middle, low);
        }
        else
        {
            // if middle does not works, then go to lower_bound(high, middle)
            return lower_bound(high, middle);
        }
    }

    static int higher_bound(int high, int low)
    {
         if (high - low == 1)
             return low;

         int middle = (high + low) / 2;


         if (simulate(middle))
             return higher_bound(high, middle);
         else
             return higher_bound(middle, low);
    }

    // simulate(location of obstacle)
    static boolean simulate(int location)
    {
        if (location == 0)
            return false;

        int max = 0;
        int cur = 0;
        for (int i = 0; i < len; i++)
        {
            char c = inp[i];
            if (c == 'L')
            {
                if (cur - 1 > location)
                    cur --;
            }
            else 
            {
                cur ++;
            }
            if (i + 1 < len)
                max = Math.max(cur, max);
        }

        return cur != 0 && max < cur;
    }
}