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;
using ll = long long;
struct frac {
  ll n, d;
};
const int MAX = 2e5 + 10;
const int MOD = 1e9 + 7;
ll t[MAX];
ll acc[MAX];
ll pot[MAX];
frac binom;
int pos;
int good;
frac sum(frac a, frac b) {
  frac r;
  r.n = (a.n * b.d + b.n * a.d) % MOD;
  r.d = (a.d * b.d) % MOD;
  return r;
}
frac prod(frac a, frac b) {
  frac r;
  r.n = (a.n * b.n) % MOD;
  r.d = (a.d * b.d) % MOD;
  return r;
}
ll red(ll x) {
  if (x < 0) return x + MOD;
  return x;
}
void decgood() {
  binom.n = (binom.n * good) % MOD;
  binom.d = (binom.d * red(pos - good + 1)) % MOD;
}
void decpos() {
  binom.d = (binom.d * pos) % MOD;
  binom.n = (binom.n * red(pos - good)) % MOD;
}
ll inv(ll x) {
  int e = MOD - 2;
  ll res = 1;
  while (e) {
    if (e % 2) res = (res * x) % MOD;
    x = (x * x) % MOD;
    e /= 2;
  }
  return res;
}
int main() {
  int n;
  ll T;
  scanf("%d%lld", &n, &T);
  pot[0] = 1;
  for (int i = 1; i <= n; i++) {
    pot[i] = (pot[i - 1] * 2) % MOD;
    scanf("%lld", &t[i]);
    acc[i] = acc[i - 1] + t[i];
  }
  binom.n = 1;
  binom.d = 1;
  pos = n;
  frac res;
  res.n = 0;
  res.d = 1;
  for (good = n; good >= 0;) {
    while (acc[pos] + (pos - good) > T) {
      if (pos == good) goto endloop;
      decpos();
      pos--;
      if (pos < 1) break;
    }
    frac temp;
    temp.n = pos;
    temp.d = pot[pos];
    if (pos < n && acc[pos + 1] + ((pos + 1) - (good + 1)) <= T)
      temp.d = pot[pos + 1];
    res = sum(res, prod(binom, temp));
  endloop:
    decgood();
    good--;
  }
  printf("%lld\n", (res.n * inv(res.d)) % MOD);
}
