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 N = 2e5 + 5;
const long long p = 1e9 + 7;
long long fac[N], inv[N];
inline long long mult(long long a, long long b) { return a * b % p; }
inline long long add(long long a, long long b) {
  a += b;
  if (a >= p) a -= p;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += p;
  return a;
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mult(res, a);
    b >>= 1;
    a = mult(a, a);
  }
  return res;
}
inline long long divi(long long a, long long b) {
  return mult(a, binpow(b, p - 2));
}
long long c(int a, int b) {
  if (b < 0 || a < b) return 0;
  return mult(mult(fac[a], inv[b]), inv[a - b]);
}
void precalc() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = mult(fac[i - 1], i);
  inv[N - 1] = binpow(fac[N - 1], p - 2);
  for (int i = N - 1; i >= 1; i--) inv[i - 1] = mult(inv[i], i);
}
long long a[N];
long long P[N];
long long pref[N];
long long two[N];
long long itwo[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  precalc();
  two[0] = 1;
  for (int i = 1; i < N; i++) {
    two[i] = mult(2, two[i - 1]);
    itwo[i] = divi(1, two[i]);
  }
  int n;
  long long T;
  cin >> n >> T;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < N; i++) {
    pref[i] = pref[i - 1] + a[i];
  }
  if (a[1] + 1 <= T)
    P[0] = 0;
  else if (a[1] + 1 == T + 1)
    P[0] = itwo[1];
  else
    P[0] = 1;
  for (int t = 1; t <= n - 1; t++) {
    long long l = max(0ll, T - pref[t + 1] + 1);
    long long r = min(t * 1ll, T - pref[t]);
    long long cur = 0;
    for (long long i = l; i <= r; i++) {
      cur = add(cur, mult(itwo[t], c(t, i)));
    }
    P[t] = mult(cur, itwo[1]);
    if (0 <= l - 1 && l - 1 <= t) cur = add(cur, mult(itwo[t], c(t, l - 1)));
    P[t] = add(P[t], mult(cur, itwo[1]));
  }
  P[n] = 1;
  for (int i = 0; i < n; i++) P[n] = sub(P[n], P[i]);
  long long ans = 0;
  for (int i = 0; i <= n; i++) ans = add(ans, mult(i, P[i]));
  cout << ans << "\n";
  return 0;
}
