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 = 2e5 + 5;
const int mod = 1e9 + 7;
long long int fast_pow(long long int a, long long int b, int modd = mod) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % modd;
    a = (a * a) % modd;
    b /= 2;
  }
  return ans;
}
vector<long long int> f(N, 1), f_1(N, 1);
long long int nCr(int n, int r) {
  long long int den = (f[r] * f[n - r]) % (mod);
  return (f[n] * fast_pow(den, mod - 2)) % mod;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  for (int j = 1; j < N; j++)
    f[j] = (f[j - 1] * j) % mod, f_1[j] = (f_1[j - 1] * j) % (mod - 1);
  long long int n, t;
  cin >> n >> t;
  vector<long long int> pre(n + 1, 0);
  vector<long long int> v(n + 1, 0);
  for (int j = 1; j <= n; j++) cin >> v[j], pre[j] = (pre[j - 1] + v[j]);
  int mn = 0, mx = 0;
  while (mn < n && (pre[mn + 1] + mn + 1) <= t) mn++;
  while (mx < n && pre[mx + 1] <= t) mx++;
  if (mn == 0) {
    cout << fast_pow(2, mod - 2) << endl;
    return 0;
  }
  long long int comp = mn;
  vector<long long int> num;
  int x = 1;
  long long int prv = 2;
  for (int j = 2; j <= mn; j++)
    prv = (((prv - 2) * 2) % mod + 2 + 2) % mod, x++;
  for (int j = mn + 1; j <= mx; j++) {
    long long int can = t - v[j];
    can -= pre[j - 1];
    assert(can >= 0);
    if (can > j) can = j;
    long long int last = nCr(j - 1, x);
    long long int upto_x =
        (((prv - 1 - last + mod + mod) * 2) % mod + (1 + last) + 1) % mod;
    if (j == 2) upto_x = 4, x = 2;
    upto_x = (upto_x + nCr(j, ++x)) % mod;
    while (can < x) {
      upto_x = (upto_x - nCr(j, x) + mod) % mod;
      x--;
    }
    prv = upto_x;
    num.push_back((upto_x * fast_pow(2, n - j)) % mod);
  }
  long long int ans = 0;
  for (auto i : num) ans = (ans + i) % mod;
  ans = (ans + (fast_pow(2, n) * mn) % mod) % mod;
  long long int den = fast_pow(2, n, mod);
  cout << (ans * fast_pow(den, mod - 2)) % mod << endl;
  return 0;
}
