# ABC281Dと近い
# 2次元dp
# dp[i][j] i番目まで見て、j mod 10、の最大枚数

N = int(input())
A = list(map(int, input().split()))

INF = 10**10
dp = [[-INF]*10 for i in range(N+1)]
dp[0][0] = 0
 
for i in range(1, N+1):
    num = A[i-1]
    for j in range(10):
        # not using i-th
        dp[i][j] = max(dp[i][j], dp[i-1][j])
        # using i-th
        dp[i][(j+num)%10] = max(dp[i][(j+num)%10], dp[i-1][j]+1)
    #print(dp[i])
    
ans = dp[N][0]
print(ans)