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 maxn = 2e5;
int N, M, C;
int n;
long long a[maxn + 5], T;
int fac[maxn + 5], ivf[maxn + 5], pw[maxn + 5];
const int mod = 1e9 + 7;
int add(int x, int y) { return (x += y) >= mod ? x - mod : x; }
int fpw(int x, int k, int p) {
  int ret = 1;
  while (k) {
    if (k & 1) ret = 1ll * ret * x % p;
    x = 1ll * x * x % p;
    k >>= 1;
  }
  return ret;
}
void prework() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ivf[n] = fpw(fac[n], mod - 2, mod);
  for (int i = n - 1; i >= 0; i--) ivf[i] = 1ll * ivf[i + 1] * (i + 1) % mod;
}
int binom(int n, int m) {
  if (n < m) return 0;
  return 1ll * fac[n] * ivf[m] % mod * ivf[n - m] % mod;
}
void moven() {
  C = add(2ll * C % mod, mod - binom(N, M));
  N++;
}
void movem1() {
  C = add(C, mod - binom(N, M));
  M--;
}
void movem2() {
  C = add(C, binom(N, M + 1));
  M++;
}
int main() {
  scanf("%d%lld", &n, &T);
  prework();
  int inv2 = (mod + 1) / 2;
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * inv2 % mod;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  if (T < a[1]) {
    puts("0");
    return 0;
  }
  N = 1, M = min(1ll, T - a[1]);
  for (int j = 0; j <= M; j++) C = add(C, binom(N, j));
  int ans = 1ll * pw[1] * C % mod;
  T -= a[1];
  for (int i = 2; i <= n; i++) {
    T -= a[i];
    if (T < 0) break;
    int curM;
    if (T > 1e9)
      curM = i;
    else
      curM = min(i, (int)T);
    moven();
    while (M < curM) movem2();
    while (M > curM) movem1();
    ans = add(ans, 1ll * pw[i] * C % mod);
  }
  cout << ans << '\n';
  return 0;
}
