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 MOD = 1e9 + 7;
inline int add(int x, int y) {
  int res = x + y;
  while (res >= MOD) res -= MOD;
  return res;
}
inline int mul(int x, int y) { return (1LL * x * y) % MOD; }
int modpow(int x, int p) {
  int res = 1;
  while (p) {
    if (p & 1) res = mul(res, x);
    x = mul(x, x);
    p /= 2;
  }
  return res;
}
inline int invert(int x) { return modpow(x, MOD - 2); }
const int N = 2e5 + 5;
int fac[N], inv[N];
void init(int n) {
  fac[0] = inv[0] = 1;
  for (int i = 1; i < n; ++i) {
    fac[i] = mul(fac[i - 1], i);
    inv[i] = invert(fac[i]);
  }
}
int comb(int n, int k) { return mul(fac[n], mul(inv[k], inv[n - k])); }
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  init(N);
  int n;
  long long T;
  cin >> n >> T;
  vector<int> v(n + 1);
  for (int i = 0; i < n; ++i) cin >> v[i];
  v[n] = T + 1;
  long long sum = 0, ans = 0;
  bool hit = false;
  int inv = invert(2), dem = 1, mmax = 0;
  for (int i = 0; i < n + 1; ++i) {
    dem = mul(dem, inv);
    sum += v[i];
    long long mx = sum + (i + 1);
    if (mx > T) {
      hit = true;
      long long lim = max(T - sum, -1LL);
      int p = 0;
      for (int j = mmax; j > lim; --j) {
        p = add(p, mul(comb(i + 1, j), dem));
      }
      p = add(p, mul(comb(i, mmax), dem));
      ans = add(ans, mul(i, p));
      mmax = lim;
    }
    if (!hit) mmax = i + 1;
  }
  cout << ans << '\n';
}
