As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).

On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:

  * It is not empty (that is n ≠ 0). 
  * The length of the sequence is even. 
  * First <image> charactes of the sequence are equal to "(". 
  * Last <image> charactes of the sequence are equal to ")". 



For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.

Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.

Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.

Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!

Input

The only line of the input contains a string s — the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.

Output

Output one number — the answer for the task modulo 109 + 7.

Examples

Input

)(()()


Output

6


Input

()()()


Output

7


Input

)))


Output

0

Note

In the first sample the following subsequences are possible:

  * If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())". 
  * If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()". 
  * If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()". 
  * If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()". 
  * If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()". 
  * If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()". 



The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Wolfgang Beyer
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            String s = in.next();
            int n = s.length();
            int[] openCounter = new int[n];
            int[] closeCounter = new int[n];

            int count = 0;
            for (int i = 0; i < n; i++) {
                if (s.charAt(i) == '(') {
                    count++;
                }
                openCounter[i] = count;
            }

            count = 0;
            for (int i = n - 1; i >= 0; i--) {
                if (s.charAt(i) == ')') {
                    count++;
                }
                closeCounter[i] = count;
            }

            long result = 0;
            Combinatorics comb = new Combinatorics(1000000007);
            for (int i = 0; i < n - 1; i++) {
                if (s.charAt(i) == '(') {
                    // long current = Combinatorics.modBinomial(openCounter[i] + closeCounter[i + 1] - 1, openCounter[i], 1000000007);
                    long current = comb.modBinomialPreCalc(openCounter[i] + closeCounter[i + 1] - 1, openCounter[i]);

                    result += current;
                    result %= 1000000007;
                }
            }
            out.println(result);


        }

    }

    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }

    static class Combinatorics {
        long[] fact;
        long[] invFact;
        int max;
        int mod;

        public Combinatorics(int modul, int m) {
            //modul should be prime

            mod = modul;
            max = m;
            fact = new long[max];
            invFact = new long[max];
            fact[1] = 1;
            fact[0] = 1;
            invFact[1] = 1;
            invFact[0] = 1;
            for (int i = 2; i < max; i++) {
                fact[i] = fact[i - 1] * i % mod;
            }

            invFact[max - 1] = modInverse(fact[max - 1], mod);
            for (int i = max - 2; i >= 0; i--) {
                invFact[i] = ((i + 1) * invFact[i + 1]) % mod; //.. 1 / i! = (i + 1) * 1 / (i + 1)!
            }
        }

        public Combinatorics(int modul) {
            // modul should be prime

            this(modul, 3000000);
        }

        public static long modPow(long a, long x, long p) {
            //calculates a^x mod p in logarithmic time.
            long res = 1;
            while (x > 0) {
                if (x % 2 != 0) {
                    res = (res * a) % p;
                }
                a = (a * a) % p;
                x /= 2;
            }
            return res;
        }

        public static long modInverse(long a, long p) {
            //calculates the modular multiplicative of a mod m.
            //(assuming p is prime).
            return modPow(a, p - 2, p);
        }

        public long modBinomialPreCalc(int n, int k) {
            if (k < 0 || k > n) return 0;
            if (k == 0 || k == n) return 1;
            //if (k > n - k) k = n - k;

            return fact[n] * invFact[k] % mod * invFact[n - k] % mod;
        }

    }
}

