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 N = 200000 + 77, Mod = 1000000007;
int n, A, a[N], Fact[N], InvFact[N], Pow[N], PowI[N];
long long T;
inline int Choice(int x, int y) {
  if (x > y || x < 0) return 0;
  return (Fact[y] * 1ll * InvFact[x] % Mod * 1ll * InvFact[y - x]) % Mod;
}
int main() {
  Fact[0] = Pow[0] = 1;
  for (int i = 1; i < N; ++i)
    Fact[i] = 1ll * i * Fact[i - 1] % Mod, Pow[i] = 2 * Pow[i - 1] % Mod;
  PowI[N - 1] = 443209094;
  InvFact[N - 1] = 731536802;
  for (int i = N - 2; i >= 0; --i)
    InvFact[i] = (i + 1) * 1ll * InvFact[i + 1] % Mod,
    PowI[i] = PowI[i + 1] * 2 % Mod;
  scanf("%d %lld", &n, &T);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  long long sum = 0;
  int val = 1, x = 0, y = 0;
  for (int i = 1; i <= n; ++i) {
    sum += a[i];
    long long rem = T - sum;
    if (rem < 0) break;
    val = (2 * val - Choice(x, y)) % Mod;
    ++y;
    if (val < 0) val += Mod;
    if (rem >= i) {
      A = (A + 1) % Mod;
      continue;
    }
    while (x > rem) val = (val - Choice(x, y) + Mod) % Mod, --x;
    while (x < rem) val = (val + Choice(x + 1, y)) % Mod, ++x;
    A = (A + val * 1ll * PowI[i]) % Mod;
  }
  printf("%d", A);
  return 0;
}
