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;
long long fact[222222];
long long factinv[222222];
long long twopows[222222];
long long GetBinom(int n, int k) {
  if (k == 0) return 1;
  long long ret = fact[n];
  ret = (ret * factinv[k]) % 1000000007;
  ret = (ret * factinv[n - k]) % 1000000007;
  return ret;
}
long long GetSum(int n, long long low, long long high) {
  if (low > high) return 0;
  if (low == 0 && high == n) {
    return twopows[n];
  }
  if (low + n - high < high - low + 1) {
    long long ret = twopows[n];
    for (int k = 0; k <= low - 1; k++) {
      ret -= GetBinom(n, k);
      ret %= 1000000007;
    }
    for (int k = high + 1; k <= n; k++) {
      ret -= GetBinom(n, k);
      ret %= 1000000007;
    }
    ret %= 1000000007;
    if (ret < 0) ret = (ret + 1000000007) % 1000000007;
    return ret;
  }
  long long ret = 0;
  for (int k = low; k <= high; k++) {
    ret += GetBinom(n, k);
    ret %= 1000000007;
  }
  return ret;
}
long long modpow(long long x, long long y) {
  long long xs = x;
  long long answer = 1;
  while (y) {
    if (y & 1) {
      answer = (answer * xs) % 1000000007;
    }
    y >>= 1;
    xs = (xs * xs) % 1000000007;
  }
  return answer;
}
long long times[222222];
int main() {
  int n;
  long long T;
  cin >> n >> T;
  fact[0] = factinv[0] = 1;
  twopows[0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    factinv[i] = modpow(fact[i], 1000000007 - 2);
    twopows[i] = twopows[i - 1] * 2 % 1000000007;
  }
  for (int i = 1; i <= n; i++) {
    cin >> times[i];
  }
  times[n + 1] = T + 1;
  long long ans = 0;
  long long cumul = 0;
  for (int i = 1; i <= n; i++) {
    cumul += times[i];
    long long twopow = modpow(twopows[i + 1], 1000000007 - 2);
    long long prob = 0;
    for (int ri = 0; ri <= 1; ri++) {
      long long up = T - cumul;
      long long low = T + 1 - cumul - times[i + 1] - ri;
      low = max(low, 0ll);
      up = min(up, (long long)i);
      prob += twopow * GetSum(i, low, up);
      prob %= 1000000007;
    }
    prob %= 1000000007;
    ans += (long long)i * prob % 1000000007;
  }
  ans %= 1000000007;
  cout << ans << endl;
}
