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 MOD = 1000000007;
long long T, n;
long long ans = 0;
vector<long long> v;
long long fact[200009];
long long factinv[200009];
long long hat2[200009];
long long hat2inv[200009];
long long modpow(long long a, long long pwr) {
  long long res = 1;
  while (pwr > 0) {
    if (pwr % 2) {
      res = (res * a) % MOD;
    }
    a = (a * a) % MOD;
    pwr /= 2;
  }
  return res;
}
long long nck(long long i, long long j) {
  if (j > i) return 0ll;
  long long res = fact[i];
  res = (res * factinv[j]) % MOD;
  res = (res * factinv[i - j]) % MOD;
  res = (res * hat2inv[i]) % MOD;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> T;
  v.push_back(0ll);
  for (int i = 0; i < n; i++) {
    long long szam;
    cin >> szam;
    v.push_back(v.back() + szam);
  }
  fact[0] = 1;
  factinv[0] = 1;
  hat2[0] = 1;
  hat2inv[0] = 1;
  for (long long i = 1; i <= n + 2; i++) {
    fact[i] = (i * fact[i - 1]) % MOD;
    factinv[i] = modpow(fact[i], MOD - 2);
    hat2[i] = (hat2[i - 1] * 2) % MOD;
    hat2inv[i] = modpow(hat2[i], MOD - 2);
  }
  for (long long i = 1; i <= n; i++) {
    if (v[i] > T) break;
    long long minimum = 0;
    if (i < n) {
      minimum = max(0ll, T - v[i + 1] + 1);
    }
    long long maximum = T - v[i];
    long long tmpans = 0;
    for (long long j = minimum; j <= min(i, maximum); j++) {
      tmpans = (tmpans + nck(i, j)) % MOD;
    }
    if (minimum > 0) {
      tmpans =
          (tmpans + (nck(i, minimum - 1) * modpow(2, MOD - 2)) % MOD) % MOD;
    }
    tmpans = (tmpans * i) % MOD;
    ans = (ans + tmpans) % MOD;
  }
  cout << ans << endl;
  return 0;
}
