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.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

	class ExtGCDResult {
		int x;
		int y;
	}

	static class Pair implements Comparable<Pair> {

		long x;
		int y;

		Pair(long x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			int comp1 = Long.compare(x, o.x);
			if (comp1 != 0) {
				return comp1;
			}
			return Integer.compare(y, o.y);
		}

		@Override
		public String toString() {
			return "(" + x + ", " + y + ")";
		}

	}

	int MOD = 1000 * 1000 * 1000 + 7;
	char[][] vals;
	int[] fact = new int[200000 + 2];

	private void solve() {
		initFact();
		char[] s = in.next().toCharArray();
		int n = s.length;
		int closing = 0;
		for (char c : s) {
			if (c == ')') {
				closing++;
			}
		}

		int opening = 0;
		int all = 0;
		for (int i = 0; i < n; i++) {
			char c = s[i];
			if (c == ')') {
				closing--;
				continue;
			}

			opening++;
			int add = C(opening + closing - 1, opening);
			all = addMod(add, all);
		}

		out.print(addMod(all, MOD));

	}

	/*
	 * 
	* 
	*/
	// --------------------------------------------------------

	private int addMod(int add, long all) {
		return (int) (((long) (add) + all) % MOD);
	}

	private int extGCD(int a, int b, ExtGCDResult res) {
		if (a == 0) {
			res.x = 0;
			res.y = 1;
			return b;
		}

		ExtGCDResult ans = new ExtGCDResult();
		int gcd = extGCD(b % a, a, ans);
		res.x = ans.y - (b / a) * ans.x;
		res.y = ans.x;
		return gcd;
	}

	private void initFact() {
		fact[0] = 1;
		for (int i = 1; i < fact.length; i++) {
			fact[i] = (int) ((((long) i) * fact[i - 1]) % MOD);
		}
	}

	private int C(int n, int k) {
		if (n < k) {
			return 0;
		}

		return divMod(fact[n], multMod(fact[k], fact[n - k]));
	}

	private int divMod(int a, int b) {
		ExtGCDResult res = new ExtGCDResult();
		int gcd = extGCD(b, MOD, res);
		int bInverse = (res.x % MOD + MOD) % MOD;
		return multMod(a, bInverse);
	}

	private int multMod(int a, int b) {
		return (int) (((long) a * b) % MOD);
	}

	private FastScanner in;
	private PrintWriter out;
	private int[][] G;

	void runIO() {

		in = new FastScanner(System.in);
		out = new PrintWriter(System.out);

		solve();

		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		}

		boolean hasMoreTokens() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

	public static void main(String[] args) {
		new Main().runIO();
	}
}