E. Devu and Flowers

E. Devu and Flowers

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains f_i 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.

Input

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, ... f_n (0 ≤ f_i ≤ 10¹²).

Output

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

Examples

Input

2 3
1 3

Output

2

Input

2 4
2 2

Output

1

Input

3 5
1 3 2

Output

3

Note

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}.

 

这个题是就是容斥原理的东西， 因为观察到N的范围就是20，S的范围就是就是很大的，所以要是在f[i]没有时间限制的情况的时候

然后就是根据常用的隔板法的原理，然后就是可以得到 x1+x2+x3+x4-----xn =s;

然后的话因为x1---xn的取值有可能是0，那么的话我得到的结果就和隔板法的结果不一样了，那这样的哈我就把所有的值都加1，然后就可以了。

还有的话我们就会用到容斥原理，然后就是假设

|!A1n!A2.....n!An| = N-|A1UA2UA3....UAn| ====》 我们假设!A== 1<=x<=f[i]+1;  那么A的话就是  f[i]+1

这样子的话区间的长度就是 sum-f[i]-1  就是从1---sum-f[i]-1;

然后的话应该就可以了，，，，，

 

 

 

转载于:https://www.cnblogs.com/Heilce/p/6402554.html