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 N = 2e5 + 7;
int qpow(int x, int y) {
  if (x == 0) return 0;
  int res = 1;
  for (; y; x = 1ll * x * x % mod, y >>= 1)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
int ny(int x) { return qpow(x, mod - 2); }
long long T;
int n, t[N], tpow[N], jc[N], njc[N], ans, wz = -1, now;
int C(int x, int y) { return 1ll * jc[x] * njc[y] % mod * njc[x - y] % mod; }
int main() {
  scanf("%d%lld", &n, &T);
  tpow[0] = jc[0] = njc[0] = 1;
  for (int i = (1), iE = (n); i <= iE; i++) tpow[i] = 2 * tpow[i - 1] % mod;
  for (int i = (1), iE = (n); i <= iE; i++)
    jc[i] = 1ll * jc[i - 1] * i % mod, njc[i] = ny(jc[i]);
  for (int i = (1), iE = (n); i <= iE; i++) scanf("%d", &t[i]);
  for (int i = (1), iE = (n); i <= iE; i++) {
    T -= t[i];
    if (T < 0)
      break;
    else if (T > i)
      ans++;
    else {
      if (wz == -1)
        wz = i, now = tpow[i];
      else
        now = (now * 2 % mod + mod - C(i - 1, wz)) % mod;
      while (wz != T) (now += mod - C(i, wz)) %= mod, wz--;
      (ans += 1ll * now * ny(qpow(2, i)) % mod) %= mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}
