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 INF = 2e18;
const int INFi = 2e9;
const int maxN = 2e5 + 5;
const int P = 998244353;
const int md = 1e9 + 7;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
int mul(int a, int b) { return (1ll * a * b) % md; }
int add(int a, int b) {
  if (a + b >= md) return a + b - md;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + md;
  return a - b;
}
int binpow(int a, int b) {
  if (b == 0) return 1;
  if (b % 2) return mul(a, binpow(a, b - 1));
  int mid = binpow(a, b / 2);
  return mul(mid, mid);
}
int rev(int a) { return binpow(a, md - 2); }
int fact[maxN];
int C(int n, int k) {
  return mul(fact[n], mul(rev(fact[k]), rev(fact[n - k])));
}
void solve() {
  fact[0] = 1;
  for (int i = 1; i < maxN; ++i) fact[i] = mul(fact[i - 1], i);
  int n;
  long long T;
  cin >> n >> T;
  vector<long long> t(n);
  for (int i = 0; (i) < n; ++(i)) cin >> t[i];
  for (int i = 1; i < n; ++i) t[i] += t[i - 1];
  int p = 0;
  int j = 0;
  for (int i = n - 1; i >= 0; --i) {
    int c = i + 1;
    j = max(0, min(j, c));
    long long time = t[i] + j;
    while (time <= T && j <= c) {
      if (c == n || t[c] + j > T) {
        int k = mul(C(c, j), binpow(2, n - c));
        p = add(p, mul(c, k));
      } else if (t[c] + j + 1 > T) {
        int k = mul(C(c, j), binpow(2, n - c - 1));
        p = add(p, mul(c, k));
      }
      j++;
      time++;
    }
    j--;
  }
  cout << mul(p, rev(binpow(2, n)));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tests = 1;
  for (int _ = 0; (_) < tests; ++(_)) {
    solve();
  }
  return 0;
}
