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;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& p) {
  return out << "(" << p.first << ", " << p.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
  for (int i = int(0); i < int(int((v).size())); i++) {
    if (i) out << " ";
    out << v[i];
  }
  return out;
}
const int MOD = int(1e9) + 7;
inline int norm(int a) {
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return int(a * 1ll * b % MOD); }
inline int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
inline int inv(int a) { return binPow(a, MOD - 2); }
int n;
long long T;
vector<long long> t;
inline bool read() {
  if (!(cin >> n >> T)) return false;
  t.resize(n);
  for (int i = int(0); i < int(n); i++) cin >> t[i];
  return true;
}
vector<int> f, inf;
int C(int n, long long k) {
  if (k > n || k < 0) return 0;
  return mul(f[n], mul(inf[k], inf[n - k]));
}
inline void solve() {
  f.resize(n + 5);
  inf.resize(n + 5);
  f[0] = inf[0] = 1;
  for (int i = int(1); i < int(int((f).size())); i++) {
    f[i] = mul(i, f[i - 1]);
    inf[i] = mul(inf[i - 1], inv(i));
  }
  int sumC = 1;
  long long bd = T;
  int pw2 = 1, i2 = (MOD + 1) / 2;
  vector<int> p(n + 1, 0);
  for (int i = int(0); i < int(n); i++) {
    pw2 = mul(pw2, i2);
    sumC = norm(mul(2, sumC) - C(i, bd));
    long long need = t[i];
    if (bd > i + 1) {
      long long sub = min(bd - i - 1, need);
      bd -= sub, need -= sub;
    }
    long long rem = min(bd + 1, need);
    for (int k = int(0); k < int(rem); k++)
      sumC = norm(sumC - C(i + 1, bd - k));
    bd -= rem;
    p[i] = mul(sumC, pw2);
  }
  int ans = int(accumulate(p.begin(), p.end(), 0ll) % MOD);
  cout << ans << endl;
}
int main() {
  cout << fixed << setprecision(15);
  if (read()) {
    solve();
  }
  return 0;
}
