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.
mod = 10 ** 9 + 7
MAX = 2 * 10 ** 5+2

    
n, T = list(map(int, input().split()))
t = [0]
t += list(map(int, input().split()))
t += [MAX]
 
r = [1] * MAX  
factorial = [1] * MAX
rfactorial = [1] * MAX
rp = [1] * MAX

sim = 0
sim_n = 0
sim_k = 0
E=0

S = [t[0]]*len(t)
for i in range(1,len(t)):
    S[i] = S[i-1]+t[i]

#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


#Permite calcular las Combinaciones de n en k.
def Combination(n,k):
    if n < k:
        return 0
    return factorial[n]*rfactorial[k]*rfactorial[n-k]


def simC(x):
    aux = 0
    mi = min(x,T-S[x])
    for i in range(0,mi+1):
        aux += Combination(x,i)%mod
    return x,mi,aux

def next_simC( next_x):
    mi1 = min(next_x,T-S[next_x])
    aux_sim = 2*sim % mod
    aux_sim = (aux_sim + Combination(sim_n,sim_k + 1))%mod
    aux_sim_n = sim_n + 1
    aux_sim_k = sim_k + 1
    while aux_sim_k > mi1:
        aux_sim = (aux_sim - Combination(aux_sim_n,aux_sim_k)) % mod
        aux_sim_k -= 1
    return aux_sim_n, aux_sim_k, aux_sim
    


 


i = 1
while i <= n and S[i] + i <= T:
    i +=1
    E +=1


if i <= n:
    sim_n,sim_k,sim = simC(i)
    
    for x in range(i,n+1):
        E = (E + rp[x]*sim)%mod

        if T-S[x+1] < 0:
            break
        
        sim_n,sim_k,sim = next_simC(x+1)
        
    
print(E)
