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;
inline long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct modint {
  long long x;
  modint(long long o = 0) { x = o; }
  modint &operator=(long long o) { return x = o, *this; }
  modint &operator+=(modint o) {
    return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
  }
  modint &operator-=(modint o) {
    return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
  }
  modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
  modint &operator^=(long long b) {
    modint a = *this, c = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) c *= a;
    return x = c.x, *this;
  }
  modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
  modint &operator+=(long long o) {
    return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
  }
  modint &operator-=(long long o) {
    return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
  }
  modint &operator*=(long long o) {
    return x = 1ll * x * o % 1000000007, *this;
  }
  modint &operator/=(long long o) {
    return *this *= ((modint(o)) ^= 1000000007 - 2);
  }
  template <class I>
  friend modint operator+(modint a, I b) {
    return a += b;
  }
  template <class I>
  friend modint operator-(modint a, I b) {
    return a -= b;
  }
  template <class I>
  friend modint operator*(modint a, I b) {
    return a *= b;
  }
  template <class I>
  friend modint operator/(modint a, I b) {
    return a /= b;
  }
  friend modint operator^(modint a, long long b) { return a ^= b; }
  friend bool operator==(modint a, long long b) { return a.x == b; }
  friend bool operator!=(modint a, long long b) { return a.x != b; }
  bool operator!() { return !x; }
  modint operator-() { return x ? 1000000007 - x : 0; }
};
modint fac[1000005], inv[1000005];
void prework(long long n) {
  fac[0] = 1;
  for (register long long i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i;
  inv[n] = modint(1) / fac[n];
  for (register long long i = (n - 1); i >= (0); --i)
    inv[i] = inv[i + 1] * (i + 1);
}
inline modint C(long long n, long long m) {
  return m > n ? modint(0) : fac[n] * inv[m] * inv[n - m];
}
long long n, t, m, a[1000005];
modint inv2 = 1, res, val = 1;
signed main() {
  prework(1000000);
  n = read(), t = read();
  for (register long long i = (1); i <= (n); ++i) a[i] = a[i - 1] + read();
  m = n;
  for (register long long i = (1); i <= (n); ++i) {
    inv2 /= 2;
    long long s = min(n, t - a[i]);
    if (s < 0) continue;
    val = val * 2 - C(i - 1, m);
    while (m > s) val -= C(i, m), m--;
    res += val * inv2;
  }
  cout << res.x;
  return 0;
}
