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  
    
25
35
43
38
33
47
44
40
36
41
42
33
30
49
42
62
39
40
35
43
31
42
46
42
34
33
 47
20
21
16
18
24
20
25
13
20
22
26
24
17
18
21
22
21
23
17
15
24
19
18
21
20
19
26
25
20
17
17
17
26
32
20
21
25
28
24
21
21
17
28
20
20
31
19
 3
343
317
337
 1
5
 10
100
100
100
100
100
100
100
100
100
100
 1
1
 13
67
75
76
80
69
86
75
86
81
84
73
72
76
 5
10
10
10
10
10
 11
291
381
126
39
19
20
3
1
20
45
2

362689152
 295545118
 691446102
 1
 12520708
 1
 232242896
 425711769
 902382672

"""
    Author - Satwik Tiwari .
    4th Oct , 2020  - Sunday
"""

#===============================================================================================
#importing some useful libraries.


from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil

from copy import deepcopy
from collections import deque


# from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
# from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
# from itertools import permutations as permutate


from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

mod = 10**9+7
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for pp in range(t):
        solve(pp)
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def nCr_mod(n,r,max_n=2 * 10**5, mod=10**9 + 7):
    max_n = min(max_n, mod - 1)

    fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
    fact[0] = 1
    for i in range(max_n):
        fact[i + 1] = fact[i] * (i + 1) % mod

    inv_fact[-1] = pow(fact[-1], mod - 2, mod)
    for i in reversed(range(max_n)):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod

    def do_nCr_mod(n, r):
        res = 1
        while n or r:
            a, b = n % mod, r % mod
            if a < b:
                return 0
            res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
            n //= mod
            r //= mod
        return res

    return do_nCr_mod(n,r)

"""
    print nCr(4,3)
"""


def solve(case):
    n = int(inp())
    ans = 1
    total = 0
    for i in range(n):
        x = int(inp())
        total+= x
        ans*=ncr(total-1,x-1)%mod

    print(ans)




testcase(1)
# testcase(int(inp()))












