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 ll = long long;
const int C = 200001, M = 1000000007;
inline ll pow(ll a, ll b, ll M) {
  if (b == 0) return 1;
  ll c = 1;
  while (b > 0) {
    if (b % 2 == 1) c = (c * a) % M;
    b /= 2;
    a = (a * a) % M;
  }
  return c;
}
ll fac[C], af[C], a[C], ipw[C];
inline ll cn(ll n, ll k) {
  if (n < k || n < 0 || k < 0) return 0;
  return (((fac[n] * af[k]) % M) * af[n - k]) % M;
}
int main() {
  int n, i;
  ll T, k, tmp_lower, P = 0, summa = 0, rs1, rs2, rs3, completed_part = 0,
                      addit;
  fac[0] = af[0] = ipw[0] = 1;
  ipw[1] = pow(2, M - 2, M);
  for (i = 1; i < C; i++) fac[i] = (fac[i - 1] * i) % M;
  for (i = 1; i < C; i++) af[i] = (af[i - 1] * pow(i, M - 2, M)) % M;
  for (i = 2; i < C; i++) ipw[i] = (ipw[i - 1] * ipw[1]) % M;
  scanf("%d %lld", &n, &T);
  for (i = 0; i < n; i++) scanf("%lld", &a[i]);
  k = (((T - 1) < (i)) ? (T - 1) : (i));
  for (i = 0; i < n; i++) {
    tmp_lower = 0;
    for (; summa + a[i] - 1 + k >= T && k >= 0; k--) {
      tmp_lower = (tmp_lower + cn(i, k) * ipw[i]) % M;
    }
    rs1 = (cn(i, T - summa - a[i] - 1) * ipw[i + 1]) % M;
    rs2 = (cn(i, T - summa - a[i]) * ipw[i + 1]) % M;
    rs3 = (cn(i, T - summa - a[i]) * ipw[i + 1]) % M;
    completed_part = (completed_part + tmp_lower + rs1 + rs2 + rs3) % M;
    P = (P + (tmp_lower + rs2) * i + (rs1 + rs3) * (i + 1)) % M;
    summa += a[i];
    k = (((T - summa - 1) < (k)) ? (T - summa - 1) : (k));
  }
  addit = ((1 - completed_part) * n) % M;
  if (addit < 0) addit += M;
  printf("%lld\n", (P + addit) % M);
  return 0;
}
