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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10, mo = 1e9 + 7;
int t[maxn];
long long s[maxn];
long long inv[maxn], prod[maxn], prodinv[maxn];
long long T;
long long ans;
int n;
long long C(int x, int y) {
  return prod[x] * prodinv[y] % mo * prodinv[x - y] % mo;
}
int main() {
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%d", &t[i]), s[i] = s[i - 1] + t[i];
  s[n + 1] = T + 1;
  inv[1] = prod[0] = prodinv[0] = prod[1] = prodinv[1] = 1;
  for (int i = 2; i <= n + 1; i++)
    inv[i] = (mo - mo / i) * inv[mo % i] % mo, prod[i] = prod[i - 1] * i % mo,
    prodinv[i] = prodinv[i - 1] * inv[i] % mo;
  long long invp2 = inv[2];
  for (int i = 1; i <= n; invp2 = invp2 * inv[2] % mo, i++)
    if (T - s[i + 1] <= i && T - s[i + 1] >= 0)
      ans = (ans + C(i, T - s[i + 1]) * invp2 % mo * inv[2] % mo * i % mo) % mo;
  long long res = 1;
  invp2 = inv[2];
  for (int i = 1, l = 0, r = 0; i <= n && s[i] <= T;
       invp2 = invp2 * inv[2] % mo,
           res = ((res * 2 - C(i, l) - C(i, r) + C(i + 1, l)) % mo + mo) % mo,
           i++)
    if (T - s[i + 1] + 1 <= i) {
      while (l > 0 && T - s[i + 1] + 1 < l) res = (res + C(i, --l) % mo) % mo;
      while (r < i && T - s[i] > r) res = (res + C(i, ++r) % mo) % mo;
      while (l < T - s[i + 1] + 1) res = (res - C(i, l++) % mo + mo) % mo;
      while (r > T - s[i]) res = (res - C(i, r--) % mo + mo) % mo;
      ans = (ans + res * invp2 % mo * i % mo) % mo;
    }
  printf("%lld\n", ans);
}
