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 long long mod = 1e9 + 7;
void Add(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
int n;
long long T, A[200005];
long long fac[200005], inv[200005];
long long Pow(long long x, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * x % mod;
    x = x * x % mod;
    b >>= 1;
  }
  return res;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  inv[n] = Pow(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}
int la, lb;
long long ans, ANS;
long long C(int a, int b) {
  if (a < b || b < 0) return 0;
  return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
void move(int b) {
  while (lb < b) Add(ans, C(la, ++lb));
  while (lb > b) Add(ans, -C(la, lb--));
}
void Solve() {
  ans = 1, la = lb = 0;
  long long inv2 = Pow(2, mod - 2), div = inv2;
  for (long long i = 1, ts = 0; i <= n; i++, div = div * inv2 % mod) {
    ts += A[i];
    int b = min(i, T - ts);
    if (b < 0) break;
    ans = (ans * 2 - C(la, lb)) % mod;
    la = i;
    move(b);
    Add(ANS, ans * div % mod);
  }
}
int main() {
  cin >> n >> T;
  for (int i = 1; i <= n; i++) cin >> A[i];
  init();
  Solve();
  ANS = (ANS % mod + mod) % mod;
  cout << ANS << endl;
  return 0;
}
