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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 1e9 + 7;
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    ex >>= 1ll;
    x = (x * x) % mod;
  }
  return ans;
}
const int maxn = 1e6 + 7;
const long long inf = 1e9 + 7;
long long f[maxn];
long long fi[maxn];
inline long long C(long long n, long long m) {
  if (n < 0) return 0ll;
  if (m > n) return 0;
  if (m < 0) return 0;
  if (m == 0) return 1ll;
  return (((f[n] * fi[m]) % mod) * fi[n - m]) % mod;
}
long long second[maxn];
long long Q[maxn];
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n;
  long long T;
  cin >> n >> T;
  second[0] = 0ll;
  f[0] = fi[0] = 1ll;
  for (int j = 1; j < maxn; j++) {
    f[j] = (f[j - 1] * 1ll * j) % mod;
    fi[j] = modexp(f[j], mod - 2);
  }
  for (int j = 1; j <= n; j++) {
    long long t;
    cin >> t;
    second[j] = second[j - 1] + t;
    Q[j] = min(max(-1ll, T - second[j]), 1ll * j);
  }
  long long ans = 0;
  long long prev = 0, lp = -1;
  for (int j = 1; j <= n; j++) {
    long long p = j - Q[j] - 1;
    assert(p >= lp);
    long long cur = modexp(2ll, j);
    long long su = prev * 2 + mod - C(j - 1, lp);
    su %= mod;
    for (int i = lp + 1; i <= p; i++) {
      su += C(j, i);
      su %= mod;
    }
    lp = p;
    prev = su;
    long long ca = (cur + mod - su) % mod;
    ca = (ca * modexp(2ll, mod - 1 - j)) % mod;
    ans = (ans + ca) % mod;
  }
  cout << ans << "\n";
}
