Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.

Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7).

Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.

The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).

The second line contains n space-separated integers f1, f2, … fn (0 ≤ fi ≤ 1012).

Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).

Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.

Sample 2. There is only one way of selecting 4 flowers: {2, 2}.

Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.

题解

有n个花坛，要选s支花，每个花坛有f[i]支花，同一个花坛的花颜色相同，不同花坛的花颜色不同，问说可以有多少种组合。

2^n的状态，枚举某些花坛的花超过了，剩下的用隔板法计算个数，再加个容斥原理就行了

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看来我应该写详细点

首先隔板法sum个球放进n个盒子中允许盒子为空的方案是C(sum+n-1,n-1)

但是由于这里有个限制，即f[i]，这样计数的话某些花会超出其个数

所以我们应该2^n枚举某些花超出了个数，比如确定i超出个数其它不确定的方案C(sum-f[i]-1+n-1,n-1)

即sum-=f[i]+1（因为允许箱子为空一开始要人为加入一个球）

这里再加个容斥原理+超0个的-超1个的+超2个的……

还有要用到乘法逆元求C和lucas定理

根据费马小定理a^(p-1)=1(mod p)，a为质数

a^(p-2)=a^(-1)(mod p)，那么a^(p-2)就是a在modp意义下的逆元

所以可以将C的公式上下分别得出取模后的乘积s1,s2，将s1乘s2的逆元

lucas定理：C(n,m)%p=C(n/p,m/p)*C(n%p,m%p)

前半部分继续递归用lucas定理，后半部分直接求