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 N = 200010, Mod = 1000000007;
long long n, T, M, res, ans;
long long a[N], inv[N], fac[N];
long long Inv[N], pow_2[N];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * inv[m] % Mod * inv[n - m] % Mod;
}
inline long long Power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power & 1) result = result * base % Mod;
    power >>= 1;
    base = (base * base) % Mod;
  }
  return result;
}
inline void Pre() {
  fac[0] = inv[0] = pow_2[0] = Inv[0] = 1;
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) a[i] += a[i - 1];
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % Mod;
  for (long long i = 1; i <= n; i++) inv[i] = Power(fac[i], Mod - 2);
  for (long long i = 1; i <= n; i++) {
    pow_2[i] = pow_2[i - 1] * 2 % Mod;
    Inv[i] = Power(pow_2[i], Mod - 2);
  }
}
signed main() {
  n = read(), T = read();
  Pre();
  M = n, res = 1;
  for (long long i = 1; i <= n; i++) {
    long long s = min(n, T - a[i]);
    if (s < 0) continue;
    res = ((2 * res - C(i - 1, M)) % Mod + Mod) % Mod;
    while (M > s) {
      res = ((res - C(i, M)) % Mod + Mod) % Mod;
      M--;
    }
    ans = (ans + res * Inv[i] % Mod) % Mod;
  }
  printf("%lld\n", ans % Mod);
  return 0;
}
