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 int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
long long t[maxn];
long long T;
int n;
int mul(int a, int b) {
  long long t = 1ll * a * b;
  return t - t / mod * mod;
}
int add(int a, int b) {
  a += b;
  if (a >= mod) return a - mod;
  return a;
}
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int fac[maxn], facinv[maxn], pw[maxn], powinv[maxn];
int C(int a, int b) {
  if (a < b) return 0;
  return mul(fac[a], mul(facinv[b], facinv[a - b]));
}
int main() {
  fac[0] = pw[0] = 1;
  for (int i = 1; i < maxn; ++i) fac[i] = mul(fac[i - 1], i);
  for (int i = 1; i < maxn; ++i) pw[i] = mul(pw[i - 1], 2);
  facinv[maxn - 1] = qp(fac[maxn - 1], mod - 2);
  powinv[maxn - 1] = qp(pw[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; --i) facinv[i] = mul(facinv[i + 1], i + 1);
  for (int i = maxn - 2; i >= 0; --i) powinv[i] = mul(powinv[i + 1], 2);
  ios::sync_with_stdio(false);
  cin >> n >> T;
  for (int i = 1; i <= n; ++i) cin >> t[i];
  for (int i = 1; i <= n; ++i) t[i] += t[i - 1];
  int ans = 0, d;
  if (t[1] > T) {
    cout << 0 << endl;
    return 0;
  }
  for (int b = 0, k; b <= n; ++b) {
    k = upper_bound(t, t + n + 1, T - b) - t - 1;
    if (k < 0) continue;
    ans = add(ans, mul(mul(C(k, b), powinv[k]), k));
    if (k != n && b != 0 && T - (t[k] + b - 1) >= t[k + 1] - t[k])
      ans = add(ans, mul(mul(C(k, b - 1), powinv[k + 1]), k));
  }
  cout << ans << endl;
  return 0;
}
