"""

表が出る確率が大きければ、期待値はひっくり返る…

表の枚数の期待値は、表の出る確率の和
確率の和 mod 200 で解けばよい

dp[i][x] = i番目まで見て、確率mod200が～の場合の数

"""

import sys
from sys import stdin

mod = 10**9+7

N = int(stdin.readline())

P = list(map(int,stdin.readline().split()))

dp = [0] * 200
dp[0] = 1

for p in P:

    ndp = [dp[j] % mod for j in range(200)]

    for j in range(200):
        ndp[(j+p)%200] += dp[j]

    dp = ndp


dp[0] -= 1

print (sum(dp[51:150]) % mod)