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;
long long modInverse(long long a, long long b) {
  return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
long long fastPow(long long a, long long pw, long long mod) {
  long long res = 1;
  a %= mod;
  while (pw > 0) {
    if (pw & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    pw >>= 1;
  }
  return res;
}
struct NchooseK {
  long long range, mod;
  vector<long long> fact, ifact;
  NchooseK(long long currMod) {
    this->mod = currMod;
    range = (long long)2e5 + 3;
    fact.resize(range + 1);
    ifact.resize(range + 1);
    calcFacts();
  }
  void calcFacts() {
    fact[0] = 1;
    for (long long i = 1; i <= range; ++i) {
      fact[i] = (1LL * fact[i - 1] * i) % this->mod;
    }
    ifact[range] = fastPow(fact[range], this->mod - 2, this->mod);
    for (long long i = range - 1; i >= 0; --i) {
      ifact[i] = (1LL * ifact[i + 1] * (i + 1)) % this->mod;
    }
  }
  long long choose(long long n, long long k) const {
    if (k < 0 || k > n || n < 0) return 0;
    return ((1LL * fact[n] * ifact[k]) % this->mod * 1LL * ifact[n - k]) %
           this->mod;
  }
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, T;
  cin >> n >> T;
  vector<long long> arr;
  long long sum = 0;
  for (long long i = 0; i < n; ++i) {
    long long val;
    cin >> val;
    sum += val;
    if (sum <= T) {
      arr.push_back(val);
    }
  }
  NchooseK nk(mod);
  n = arr.size();
  vector<long long> powHalf(n + 2, 1);
  const long long half = fastPow(2, mod - 2, mod);
  assert((half + half) % mod == 1);
  for (long long i = 1; i < (long long)powHalf.size(); ++i) {
    powHalf[i] = (powHalf[i - 1] * half) % mod;
  }
  sum = 0;
  long long res = 0;
  long long prefixPascal = 1;
  long long lastBot = 0;
  for (long long i = 0; i < n; ++i) {
    sum += arr[i];
    if (sum + i + 1 <= T) {
      ++res;
    } else {
      assert(sum <= T);
      long long X = min(T - sum, i + 1);
      prefixPascal *= 2;
      prefixPascal -= nk.choose(i, lastBot);
      prefixPascal %= mod;
      prefixPascal += mod;
      prefixPascal %= mod;
      while (lastBot < X) {
        ++lastBot;
        prefixPascal += nk.choose(i + 1, lastBot);
        prefixPascal %= mod;
      }
      while (lastBot > X) {
        prefixPascal -= nk.choose(i + 1, lastBot) + 2 * mod;
        prefixPascal %= mod;
        --lastBot;
      }
      res += prefixPascal * powHalf[i + 1] % mod;
      res %= mod;
      res += mod;
      res %= mod;
    }
  }
  cout << res << '\n';
}
