def solve(arr, target):
    # python 0/1 背包仿 bitset 解
    # python 0/1 Knapsack Problem like bitset
    # 此解求出 大於等於 目標子集和的解
    a = 1
    for x in arr: a |= (a << x)
    a = bin(a)[2:]
    return a.find('1', target)

def main():
    from sys import stdin
    e = stdin.readline
    # m 是櫃子總數 s 是需求
    m, s, n = map(int, e().split())
    # sum(f) 是被借出的合計
    f = [int(x) for x in e().split()]
    # b 是剩餘
    b = m - sum(f)
    if(b >= s): print(0)
    else:
       # s - b 是真正的需求 sb
       # 還回的量需 >= sb
       r = solve(f, s - b)
       print(r)
main()