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;
template <class T>
bool tomin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <class T>
bool tomax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
void read(T &x) {
  char c;
  x = 0;
  int f = 1;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') f = -1;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while (c = getchar(), c >= '0' && c <= '9');
  x *= f;
}
const double Pi = acos(-1);
const int maxn = 2e5 + 5;
const int P = 1e9 + 7;
int n;
long long T, sum[maxn];
long long fact[maxn], finv[maxn], pw[maxn];
long long Com(int n, int m) {
  if (m < n) return 0;
  return fact[m] * finv[m - n] % P * finv[n] % P;
}
void Check(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int main() {
  srand(time(NULL));
  read(n), read(T);
  fact[0] = finv[0] = finv[1] = pw[0] = 1;
  for (int i = 1, i_ = n + 1; i <= i_; ++i) fact[i] = fact[i - 1] * i % P;
  for (int i = 2, i_ = n + 1; i <= i_; ++i)
    finv[i] = P - P / i * finv[P % i] % P;
  for (int i = 1, i_ = n + 1; i <= i_; ++i) finv[i] = finv[i - 1] * finv[i] % P;
  for (int i = 1, i_ = n + 1; i <= i_; ++i) pw[i] = pw[i - 1] * finv[2] % P;
  for (int i = 1, i_ = n; i <= i_; ++i) {
    read(sum[i]);
    sum[i] += sum[i - 1];
  }
  int ans = 0;
  long long L, R;
  for (int i = 1, i_ = n - 1; i <= i_; ++i) {
    if (sum[i] > T) break;
    R = min(T - sum[i], (long long)i);
    int sumc = 0;
    L = max((long long)0, T - sum[i + 1] + 1);
    if (L <= R) {
      for (int j = L, j_ = R; j <= j_; ++j) Check(sumc, Com(j, i));
      Check(ans, pw[i + 1] * sumc % P * i % P);
    }
    L = max((long long)0, T - sum[i + 1]);
    if (L <= R) {
      if (T >= sum[i + 1]) Check(sumc, Com(L, i));
      Check(ans, pw[i + 1] * sumc % P * i % P);
    }
  }
  L = 0, R = min(T - sum[n], (long long)n);
  int sumc = 0;
  for (int i = L, i_ = R; i <= i_; ++i) Check(sumc, Com(i, n));
  if (L <= R) Check(ans, pw[n] * sumc % P * n % P);
  printf("%d", ans);
  return 0;
}
