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

import sys
from collections import defaultdict, Counter
from itertools import permutations, combinations
from math import sin, cos, asin, acos, tan, atan, pi

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition, yes = "YES", no = "NO", none = "-1") :
  if condition == None:
    print (none)
  elif condition :
    print (yes)
  else :
    print (no)

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rstr() :
  return sys.stdin.readline().strip()


def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits[::-1]

def undigits(x, p) :
  value = 0
  for d in x :
    value *= p
    value += d
  return value

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def check_rectangle(points) :
  assert(len(points) == 4)

  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    sides = (
      vector_diff(A1, A2),
      vector_diff(A2, A3),
      vector_diff(A3, A4),
      vector_diff(A4, A1),
    )
    if all(scalar_product(s1, s2) == 0 for s1, s2 in zip(sides, sides[1:])) :
       return True
  return False

def check_square(points) :
  if not check_rectangle(points) :
    return False
  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    side_lengths = [
      (first[0] - next[0]) ** 2 + (first[1] - next[1]) ** 2 for first, next in zip([A1, A2, A3, A4], [A2, A3, A4, A1])
    ]
    if len(set(side_lengths)) == 1 :
      return True
    
  return False

def check_right(p) :
  # Check if there are same points
  for a, b in [
    (p[0], p[1]),
    (p[0], p[2]),
    (p[1], p[2]),
  ] :
    if a[0] == b[0] and a[1] == b[1] :
      return False

  a, b, c = p
  a, b, c = vector_diff(a, b), vector_diff(b, c), vector_diff(c, a)   

  return scalar_product(a, b) * scalar_product(a, c) * scalar_product(b, c) == 0

def modmatrixproduct(a, b, mod) :
  n, m1 = len(a), len(a[0])
  m2, k = len(b), len(b[0])

  assert(m1 == m2)
  m = m1

  r = [[0] * k for i in range(n)]
  for i in range(n) :
    for j in range(k) :
      for l in range(m) :
        r[i][j] += a[i][l] * b[l][j]
      r[i][j] %= mod
  return r

def modmatrixpower(a, n, mod) :
  magic = 2
  for m in [2, 3, 5, 7] :
    if n % m == 0 :
      magic = m
      break

  r = None
  if n < magic : 
    r = a
    n -= 1
  else :
    s = modmatrixpower(a, n // magic, mod)
    r = s
    for i in range(magic - 1) :
      r = modmatrixproduct(r, s, mod)

  for i in range(n % magic) : 
    r = modmatrixproduct(r, a, mod)
  
  return r

k = rint()

a = [rint() for i in range(k)]

variants = 1
elements = 0

from math import factorial
for v in a :
  variants *= factorial(elements + v - 1) / factorial(elements) / factorial(v - 1)
  elements += v
  variants %= 1000000007

print variants
