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.
from sys import stdin, stdout, exit

mod = 10**9 + 7

def modinv(x):
    return pow(x, mod-2, mod)

N = 2*10**5 + 10
facts = [1]*N
for i in range(1,N):
    facts[i] = facts[i-1] * i
    facts[i] %= mod

def binom(n, k):
    ans = modinv(facts[k]) * modinv(facts[n-k])
    ans %= mod
    ans *= facts[n]
    ans %= mod
    return ans

#print("Finished preprocess")

n, T = map(int, stdin.readline().split())
ts = list(map(int, stdin.readline().split()))

ans = 0
total = sum(ts)
running = total
last_idx = n-1
while running > T:
    running -= ts[last_idx]
    last_idx -= 1
#print(last_idx+1)

last_bd = -1
last_sum = 0
idx = last_idx
while running + idx + 1 > T:
    bd = T - running
#    print("time remaining for", idx+1, "flips is", bd)
    cur_sum = last_sum + (binom(idx+1, last_bd) if last_bd >= 0 else 0)
    cur_sum *= modinv(2)
    cur_sum %= mod
    for fresh in range(last_bd+1, bd+1):
        cur_sum += binom(idx+1, fresh)
        cur_sum %= mod
 #   print("pr of", idx+1, "flips is", cur_sum, cur_sum / (2**(idx+1)))
    ans += cur_sum * modinv(pow(2, idx+1, mod))
    ans %= mod
    running -= ts[idx]
    last_bd = bd
    last_sum = cur_sum
    idx -= 1

#print(idx+1, "freebies")
ans += idx+1
ans %= mod
print(ans)
