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>
long long int const MOD = 1000000007;
long long int const N = 1000005;
long long int const LN = 20;
long long int const inf = 8e18;
using namespace std;
long long int n, m, a[N];
long long int fact[N];
long long int ifact[N];
long long int bexpo(long long int x, long long int n, long long int m) {
  long long int r = 1;
  x = x % m;
  while (n > 0) {
    if (n % 2) {
      r = (r * x) % m;
    }
    x = (x * x) % m;
    n = n / 2;
  }
  return r;
}
long long int ncr(long long int n, long long int r) {
  if (r > n) {
    return 0;
  }
  long long int ans = fact[n];
  ans = (ans * ifact[r]) % MOD;
  ans = (ans * ifact[n - r]) % MOD;
  return ans;
}
void pre() {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long int i = 1; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  ifact[N - 1] = bexpo(fact[N - 1], MOD - 2, MOD);
  for (long long int i = N - 2; i > 0; i--) {
    ifact[i] = (ifact[i + 1] * (i + 1)) % MOD;
  }
}
long long int p2[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  p2[0] = 1;
  for (long long int i = 1; i < N; i++) {
    p2[i] = 2 * p2[i - 1] % MOD;
  }
  pre();
  long long int n, t;
  cin >> n >> t;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long int sm = 0;
  long long int ans = 0;
  long long int prev = 1;
  long long int idx = 0;
  for (long long int i = 1; i <= n; i++) {
    sm += a[i];
    long long int temp = t - sm;
    if (temp < 0) {
      break;
    }
    temp = min(i, temp);
    long long int inv2 = bexpo(p2[i], MOD - 2, MOD);
    long long int ok = 2 * prev;
    if (ok < 0) {
      ok += MOD;
    }
    ok -= ncr(i - 1, idx);
    ok %= MOD;
    if (ok < 0) {
      ok += MOD;
    }
    ok %= MOD;
    while (idx > temp) {
      ok -= ncr(i, idx);
      idx--;
      ok %= MOD;
      if (ok < 0) {
        ok += MOD;
      }
    }
    while (idx < temp) {
      ok += ncr(i, idx + 1);
      idx++;
      ok %= MOD;
    }
    prev = ok;
    idx = temp;
    ok *= inv2;
    ok %= MOD;
    ans += ok;
    ans %= MOD;
  }
  cout << ans;
  return 0;
}
