Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color i before drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen. 

Input

The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.

Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).

The total number of balls doesn't exceed 1000.

Output

A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007. 

Examples

Input

3
2
2
1


Output

3


Input

4
1
2
3
4


Output

1680

Note

In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: 
    
    
      
    1 2 1 2 3  
    1 1 2 2 3  
    2 1 1 2 3  
    
import java.util.*;
import java.io.*;
import java.math.*;
	public class Main {
		public static long mod= (long) (1e9 +7);
		static long ncr[][] = new long[1000+7][1000+7];
		public static void main(String args[])
		{
			InputReader s= new InputReader(System.in);
			OutputStream outputStream= System.out;
			PrintWriter out= new PrintWriter(outputStream);
			int n= s.nextInt();
			long ans = 1;
	        int sum = 0;
	        precomputation();
	        for(int i=0;i<n;i++){
	            int temp=s.nextInt();
	            sum+=temp;
	            ans=((ans%mod)*(ncr[sum-1][temp-1]%mod))%mod;
	        }
			out.println(ans%mod);
			out.close();
		}
		static void precomputation(){
	        ncr[0][0] = 1;
	        
	        for(int i=1;i<=1000;i++){
	            for(int j=0;j<=i;j++){
	                if(j==0)
	                    ncr[i][j] = 1;
	                else{
	                    ncr[i][j] = (ncr[i-1][j-1] + ncr[i-1][j])%mod;
	                }
	            }
	        }
	    }
		static long combinations1(long n,long r){   // O(r)
			if(r==0 || r==n) return 1;
			r= Math.min(r, n-r);
			long ans=n;  // nC1=n;
			for(int i=1;i<r;i++){
				ans= ((ans%mod)*((n-i)%mod))%mod;
				ans=  (long) ((ans%mod)*(Math.pow(i+1, -1))%mod)%mod;
			}
			return ans%mod;
			
		}
		static long combinations(long n,long r){   // O(r)
			if(r==0 || r==n) return 1;
			r= Math.min(r, n-r);
			long ans=n;  // nC1=n;
			for(int i=1;i<r;i++){
				ans= ans*(n-i);
				ans= ans/(i+1);
			}
			return ans;
			
		}
		static long modulo(long a,long b,long c) {
			long x=1;
			long y=a;
			while(b > 0){
				if(b%2 == 1){
					x=(x*y)%c;
				}
				y = (y*y)%c; // squaring the base
				b /= 2;
			}
			return  x%c;
		}
		static void catalan_numbers(int n) {
			long catalan[]= new long[n+1];
		    catalan[1] = 1;
		    for (int i = 2; i <= n; i++) {
		        for (int j = 1; j <= i - 1; j++) {
		            catalan[i] = catalan[i] + ((catalan[j]) * catalan[i - j]);
		        }
		    }
		}
		
		static ArrayList<Integer> primeFactors(int n)   // O(sqrt(n))
		{
		    // Print the number of 2s that divide n
			ArrayList<Integer> arr= new ArrayList<>();
		    while (n%2 == 0)
		    {
		        arr.add(2);
		        n = n/2;
		    }
		 
		    // n must be odd at this point.  So we can skip one element (Note i = i +2)
		    for (int i = 3; i <= Math.sqrt(n); i = i+2)
		    {
		        // While i divides n, print i and divide n
		        while (n%i == 0)
		        {
		            arr.add(i);
		            n = n/i;
		        }
		    }
		    // This condition is to handle the case when n is a prime number
		    // greater than 2
		    if (n > 2)
		        arr.add(n);
		    
		    return arr;
		}
	
		public static long expo(long a, long b){
			if (b==0)
		        return 1%mod;
		    if (b==1)
		        return a%mod;
		    if (b==2)
		        return ((a%mod)*(a%mod))%mod;
	
		    if (b%2==0){
		            return expo(expo(a%mod,(b%mod)/2)%mod,2%mod)%mod;
		    }
		    else{
		        return (a%mod)*(expo(expo(a%mod,(b-1)%mod/2)%mod,2%mod)%mod)%mod;
		    }
		}
		static class Pair implements Comparable<Pair>
		{
			long f;
			String s;
			Pair(long ii, String cc)
			{
				f=ii;
				s=cc;
			}
			
			public int compareTo(Pair o) 
			{
				return Long.compare(this.f, o.f);
			}
			
		}
		
		public static int[] sieve(int N){    // O(n*log(logn))
			int arr[]= new int[N+1];
			for(int i=2;i<Math.sqrt(N);i++){
				if(arr[i]==0){
					for(int j= i*i;j<= N;j= j+i){
						arr[j]=1;
					}
				}
			}
			return arr;
			// All the i for which arr[i]==0 are prime numbers.
		}
		static long gcd(long a,long b){       // O(logn)
			if(b==0) return a;
			return gcd(b,a%b);
		}
		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());
		    }
		  }
	}