1194_F. Crossword Expert
Today Adilbek is taking his probability theory test. Unfortunately, when Adilbek arrived at the university, there had already been a long queue of students wanting to take the same test. Adilbek has estimated that he will be able to start the test only T seconds after coming. 

Fortunately, Adilbek can spend time without revising any boring theorems or formulas. He has an app on this smartphone which contains n Japanese crosswords to solve. Adilbek has decided to solve them all one by one in the order they are listed in the app, without skipping any crossword. For each crossword, a number t_i is given that represents the time it takes an average crossword expert to solve this crossword (the time is given in seconds).

Adilbek is a true crossword expert, but, unfortunately, he is sometimes unlucky in choosing the way to solve the crossword. So, it takes him either t_i seconds or t_i + 1 seconds to solve the i-th crossword, equiprobably (with probability 1/2 he solves the crossword in exactly t_i seconds, and with probability 1/2 he has to spend an additional second to finish the crossword). All these events are independent.

After T seconds pass (or after solving the last crossword, if he manages to do it in less than T seconds), Adilbek closes the app (if he finishes some crossword at the same moment, that crossword is considered solved; otherwise Adilbek does not finish solving the current crossword at all). He thinks it would be an interesting probability theory problem to calculate E — the expected number of crosswords he will be able to solve completely. Can you calculate it? 

Recall that the expected value of a discrete random variable is the probability-weighted average of all possible values — in this problem it means that the expected value of the number of solved crosswords can be calculated as E = ∑ _{i = 0}^{n} i p_i, where p_i is the probability that Adilbek will solve exactly i crosswords. 

We can represent E as rational fraction P/Q with Q > 0. To give the answer, you should print P ⋅ Q^{-1} mod (10^9 + 7).

Input

The first line contains two integers n and T (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ T ≤ 2 ⋅ 10^{14}) — the number of crosswords and the time Adilbek has to spend, respectively.

The second line contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ 10^9), where t_i is the time it takes a crossword expert to solve the i-th crossword.

Note that Adilbek solves the crosswords in the order they are given in the input without skipping any of them.

Output

Print one integer — the expected value of the number of crosswords Adilbek solves in T seconds, expressed in the form of P ⋅ Q^{-1} mod (10^9 + 7).

Examples

Input


3 5
2 2 2


Output


750000007


Input


3 5
2 1 2


Output


125000003

Note

The answer for the first sample is equal to 14/8.

The answer for the second sample is equal to 17/8.
```python
mod = 10 ** 9 + 7
MAX = 2 * 10 ** 5+2
 
r = [1] * MAX  
factorial = [1] * MAX
rfactorial = [1] * MAX
rp = [1] * MAX 

#Permite precalcular factorial hasta "MAX", para evitar tener que calcularlo varias veces
#dentro de la ejecucion del programa.
for i in range(2, MAX):
    factorial[i]   = i * factorial[i - 1] % mod
    
    r[i] = mod - (mod // i) * r[mod%i] % mod

    rfactorial[i] = rfactorial[i-1] * r[i] % mod

#Calcula el inverso de "p" para evitar usar fracciones racionales.    
for i in range(1, MAX):
    rp[i] = rp[i - 1] * (mod + 1) // 2 % mod
    
n, T = list(map(int, input().split()))
t = list(map(int, input().split()))
t.append(10**10+1)

#Permite calcular las Combinacione de n en k.
def Combination(n,k):
    return factorial[n]*rfactorial[k]*rfactorial[n-k]
 
S=0
E=0

for i in range(0,n+1):
    sim = 0
    for add in range(2):
        l_, r_ =  max(0, T-S-t[i] + add), min(i, T-S)
        
        for x in range(l_, r_+1):
            sim += Combination(i,x) * rp[i+1]    
    E = (E + i * sim) % mod
    S += t[i]   
    
print(E)

```