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.util.*;
import java.io.*;

public class CF404D 
{
	static long[] f;
	static void fact()
	{
		f=new long[(int)2e5+5];
		f[0]=1;
		for(int i=1;i<=(int)2e5+1;i++)
			f[i]=(((long)i*(long)f[i-1])%l);
	}
	public static void main(String args[]) {
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		/*------------------------------My Code starts here------------------------------*/
		String s=in.next();
		int close=0,i,n=s.length(),open=0;
		long ans=0;
		for(i=0;i<n;i++)
		{
			if(s.charAt(i)==')')
				close++;
		}
		fact();
		for(i=0;i<n;i++)
		{
			if(s.charAt(i)==')')
				close--;
			else
			{
				open++;
				if(close>0)
					ans+=NcR(open+close-1,open);
				ans%=l;
			}
		}
		out.println(ans);
		out.close();
		/*------------------------------The End------------------------------------------*/
	}

	private static long NcR(int n, int r) 
	{
		long ans=f[n]%l;
		ans*=modPow(f[n-r], l-2, l);
		ans%=l;
		ans*=modPow(f[r], l-2, l);
		ans%=l;
		return ans;
	}
	public static final long l = (int) (1e9 + 7);
	static long modPow(long a, long b, long p) {
		long result = 1;
		a %= p;
		while (b > 0) {
			if ((b & 1) != 0)
				result = (result * a) % p;
			b = b >> 1;
		a = (a * a) % p;
		}
		return result;
	}
	private static int[] nextIntArray(InputReader in, int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = in.nextInt();
		return a;
	}

	private static long[] nextLongArray(InputReader in, int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = in.nextLong();
		return a;
	}

	private static int[][] nextIntMatrix(InputReader in, int n, int m) {
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++)
				a[i][j] = in.nextInt();
		}
		return a;
	}

	private static void show(int[] a) {
		for (int i = 0; i < a.length; i++)
			System.out.print(a[i] + " ");
		System.out.println();
	}

	private static void show2DArray(char[][] a) {
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a[0].length; j++)
				System.out.print(a[i][j]);
			System.out.println();
		}
	}

	static class Pair {
		private int first;
		private int second;

		public Pair(int i, int j) {
			this.first = i;
			this.second = j;
		}

		public int getFirst() {
			return first;
		}

		public int getSecond() {
			return second;
		}

		public void setFirst(int k) {
			this.first = k;
		}

		public void setSecond(int k) {
			this.second = k;
		}
	}
	public static void SieveOfEratosthenes(int n) {
		boolean[] prime = new boolean[n + 1];
		Arrays.fill(prime, true);
		prime[1] = false;
		int i, j;
		for (i = 2; i * i <= n; i++) {
			if (prime[i]) {
				for (j = i; j <= n; j += i) {
					if (j != i)
						prime[i] = false;
				}
			}
		}
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine() {
			String fullLine = null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine = reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

}
