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>
inline long long read() {
  char c, _c;
  long long x;
  for (c = _c = getchar(); !isdigit(c); c = getchar()) {
    _c = c;
  }
  for (x = 0; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  return _c == '-' ? -x : x;
}
const int mod = 1e9 + 7, inv2 = mod + 1 >> 1;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int sub(int x, int y) { return x - y >= 0 ? x - y : x - y + mod; }
inline int power(int x, int y, int res = 1) {
  for (; y; y >>= 1, x = 1ll * x * x % mod) {
    if (y & 1) {
      res = 1ll * res * x % mod;
    }
  }
  return res;
}
const int N = 2e5 + 5;
int n, k, ans, now = 1, tot = 1, fac[N], invf[N];
long long m;
void initCombin(int n) {
  for (int i = fac[0] = 1; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  invf[n] = power(fac[n], mod - 2);
  for (int i = n; i; i--) {
    invf[i - 1] = 1ll * invf[i] * i % mod;
  }
}
int binom(int n, int m) {
  return n < m ? 0 : 1ll * fac[n] * invf[m] % mod * invf[n - m] % mod;
}
int main() {
  n = read();
  m = read();
  initCombin(k = n);
  for (int i = 0; i < n; i++) {
    m -= read();
    if (m < 0) {
      break;
    }
    for (; k > m; k--) {
      now = sub(now, binom(i, k));
    }
    now = sub(add(now, now), binom(i, k));
    tot = 1ll * tot * inv2 % mod;
    ans = add(ans, 1ll * tot * now % mod);
  }
  printf("%d\n", ans);
  return 0;
}
