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 mo = 1000000007;
template <typename _T>
_T Fabs(_T x) {
  return x < 0 ? -x : x;
}
template <typename _T>
void read(_T &x) {
  _T f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
long long n, fac[200005], inv[200005], f[200005], ans;
long long pow2[200005], inv2[200005], t[200005], T;
long long qkpow(long long a, long long s) {
  long long t = 1;
  while (s) {
    if (s & 1) t = 1ll * a * t % mo;
    a = 1ll * a * a % mo;
    s >>= 1;
  }
  return t;
}
void init() {
  inv[0] = fac[0] = f[1] = inv[1] = fac[1] = 1;
  for (int i = 2; i <= 2e5; i++) {
    fac[i] = fac[i - 1] * i % mo;
    f[i] = (mo - mo / i) * f[mo % i] % mo;
    inv[i] = inv[i - 1] * f[i] % mo;
  }
  pow2[0] = inv2[0] = 1;
  for (int i = 1; i <= 2e5; i++) {
    pow2[i] = pow2[i - 1] * 2ll % mo;
    inv2[i] = inv2[i - 1] * f[2] % mo;
  }
}
long long C(long long x, long long y) {
  if (x < 0 || y < 0 || y > x) return 0;
  return fac[x] * inv[y] % mo * inv[x - y] % mo;
}
signed main() {
  read(n);
  read(T);
  init();
  long long now = n, val = 1;
  for (int i = 1; i <= n; i++) read(t[i]), t[i] += t[i - 1];
  for (int i = 1; i <= n; i++) {
    int s = min(n, T - t[i]);
    if (s < 0) break;
    val = (2ll * val - C(i - 1, now) + mo) % mo;
    while (now > s) val = (val - C(i, now) + mo) % mo, now--;
    ans = (ans + val * inv2[i] % mo) % mo;
  }
  printf("%lld\n", (ans + mo) % mo);
  return 0;
}
