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>
const double pi = acos(-1.0);
const double eps = 1e-7;
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 3e5 + 1;
long long bin_pow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % MOD;
    n >>= 1;
    x = x * x % MOD;
  }
  return res;
}
long long add(long long x, long long y) {
  x += y;
  if (x >= MOD) x -= MOD;
  return x;
}
long long mult(long long x, long long y) { return x * y % MOD; }
long long rev_pow(long long x, long long n) { return bin_pow(x, MOD - n - 1); }
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += MOD;
  return x;
}
long long fact[N], rev_fact[N];
void fac(int ffrom) {
  fact[0] = 1;
  for (int i = 1; i <= ffrom; i++) fact[i] = mult(fact[i - 1], 1ll * i);
  rev_fact[ffrom] = rev_pow(fact[ffrom], 1);
  for (int i = ffrom - 1; i >= 0; i--)
    rev_fact[i] = mult(rev_fact[i + 1], 1ll * (i + 1));
}
long long C(int n, int k) {
  if (k > n || k < 0) return 0;
  return mult(mult(fact[n], rev_fact[k]), rev_fact[n - k]);
}
void solve() {
  int n;
  long long T;
  cin >> n >> T;
  fac(n + 10);
  vector<long long> t(n + 1);
  for (int i = 1; i <= n; i++) cin >> t[i];
  long long sum = 0;
  long long last_ans = 0;
  int last_id = -2;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    sum += t[i];
    if (sum > T) break;
    int now_id;
    long long now_ans;
    if (T - sum >= i)
      now_id = i;
    else
      now_id = T - sum;
    now_ans = add(mult(last_ans, 2), C(i - 1, last_id + 1));
    now_ans = sub(now_ans, C(i, last_id + 1));
    if (now_id > last_id) {
      for (int j = max(-1, last_id) + 1; j <= now_id; j++)
        now_ans = add(now_ans, C(i, j));
    } else if (now_id < last_id) {
      for (int j = now_id + 1; j <= last_id; j++)
        now_ans = sub(now_ans, C(i, j));
    }
    ans = add(ans, mult(now_ans, rev_pow(2, i)));
    last_ans = now_ans;
    last_id = now_id;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}
