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 = 5 * 10 ** 5
fac, ifac = [1] * MAX, [1] * MAX
for i in range(2, MAX):
    fac[i] = fac[i - 1] * i % MOD
ifac[-1] = pow(fac[-1], MOD - 2, MOD)
for i in range(MAX - 2, 1, -1):
    ifac[i] = ifac[i + 1] * (i + 1) % MOD
ipow2 = [1] * MAX
for i in range(1, MAX):
    ipow2[i] = ipow2[i - 1] * (MOD + 1) // 2 % MOD

choose = lambda n, k: fac[n] * ifac[k] % MOD * ifac[n - k] % MOD

n, t = map(int, input().split())
a = list(map(int, input().split()))

s = 0
p = [1] + [0] * (n + 1)
k = cur = 0
for i in range(n):
    s += a[i]
    if s > t: break
    if s + i + 1 <= t:
        p[i + 1] = 1
        continue
    if not cur:
        k = t - s
        for j in range(k + 1):
            cur += choose(i + 1, j)
        cur %= MOD
    else:
        cur = cur * 2 - choose(i, k)
    while k > t - s:
        cur -= choose(i + 1, k)
        k -= 1
    cur %= MOD
    p[i + 1] = cur * ipow2[i + 1] % MOD

print(sum((p[i] - p[i + 1]) * i % MOD for i in range(1, n + 1)) % MOD)