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 = 2e5 + 5;
const long long M = 1e6 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
inline long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, b, fac[N], inv[N], p[N], _n, _m, res, s[N];
inline void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (long long i = (2); i <= (N); ++i)
    fac[i] = fac[i - 1] * i % mod,
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  for (long long i = (2); i <= (N); ++i) inv[i] = inv[i] * inv[i - 1] % mod;
}
inline long long C(long long n, long long m) {
  return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
inline long long qpow(long long n, long long m) {
  if (m < 0) return 0;
  long long res = 1;
  for (; m; m >>= 1) {
    if (m & 1) res = res * n % mod;
    n = n * n % mod;
  }
  return res;
}
const long long inv2 = qpow(2, mod - 2);
inline long long S(long long n, long long m) {
  m = min(n, m);
  if (!_n) {
    for (long long i = (0); i <= (m); ++i) (res += C(n, i)) %= mod;
    _n = n, _m = m;
  } else {
    for (; _n < n; ++_n) res = (res * 2 - C(_n, _m) + mod) % mod;
    for (; _m > m; --_m) res = (res - C(_n, _m) + mod) % mod;
    for (; _m < m; ++_m) (res += C(_n, _m + 1)) %= mod;
  }
  return res;
}
signed main() {
  init();
  a = read();
  b = read();
  for (long long i = (1); i <= (a); ++i) s[i] = read();
  long long t = inv2;
  for (long long i = (1); i <= (a); ++i) {
    b -= s[i];
    if (b < 0) break;
    p[i] = S(i, b) * t % mod;
    (t *= inv2) %= mod;
  }
  long long ans = 0;
  for (long long i = (1); i <= (a); ++i) {
    (ans += (p[i] - p[i + 1] + mod) % mod * i % mod) %= mod;
  }
  printf("%lld", ans);
  return 0;
}
