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 <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ": " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = false;
  string r = "{";
  for (auto x : v) {
    if (f) r += ", ";
    r += to_string(x);
    f = true;
  }
  return r += "}";
}
void err(istream_iterator<string>) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << to_string(a) << endl;
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 1e9 + 7;
long long const INF = 1e18;
inline int add(int a, int b, int mod = MOD) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int sub(int a, int b, int mod = MOD) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline int mul(int a, int b, int mod = MOD) {
  return int((long long)a * b % mod);
}
inline int mpow(int base, long long ex, int mod = MOD) {
  int res = 1;
  while (ex > 0) {
    if (ex & 1) res = mul(res, base, mod);
    base = mul(base, base, mod);
    ex >>= 1;
  }
  return res;
}
inline int inv(int a, int mod = MOD) { return mpow(a, mod - 2, mod); }
vector<int> fac, ifac;
inline int ncr(int n, int r) {
  if (n < r) return 0;
  return mul(fac[n], mul(ifac[n - r], ifac[r]));
}
void prepare_factorial(int nax = 1e6 + 10) {
  fac.resize(nax);
  ifac.resize(nax);
  fac[0] = 1;
  for (int i = 1; i < nax; ++i) fac[i] = mul(i, fac[i - 1]);
  ifac[nax - 1] = inv(fac[nax - 1]);
  for (int i = nax - 1; i > 0; --i) ifac[i - 1] = mul(i, ifac[i]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  prepare_factorial();
  int n;
  long long t;
  cin >> n >> t;
  vector<long long> a(n);
  for (auto &z : a) cin >> z;
  for (int i = 1; i < n; ++i) {
    a[i] += a[i - 1];
  }
  while (!a.empty() && a.back() > t) a.pop_back();
  n = int((a).size());
  if (n == 0) kek(0);
  vector<int> p(n + 2);
  p[0] = 1;
  int ans = 0;
  int cur = 1, k = n;
  for (int i = 1; i <= n; ++i) {
    long long d = t - a[i - 1];
    cur = sub(add(cur, cur), ncr(i - 1, k));
    if (d >= i) {
      p[i] = 1;
    } else {
      while (k > d) cur = sub(cur, ncr(i, k--));
      while (k < d) cur = add(cur, ncr(i, ++k));
      p[i] = mul(cur, inv(mpow(2, i)));
    }
  }
  for (int i = 0; i <= n; ++i) {
    ans = add(ans, mul(i, sub(p[i], p[i + 1])));
  }
  kek(ans);
  return 0;
}
