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>
template <class C>
C& mini(C& a, C b) {
  if (b < a) a = b;
  return a;
}
template <class C>
C& maxi(C& a, C b) {
  if (a < b) a = b;
  return a;
}
using namespace std;
template <typename T>
void read(T& x) {
  x.read();
}
template <typename T>
void read(vector<T>& v) {
  for (T& x : v) read(x);
}
template <typename X, typename Y>
void read(pair<X, Y>& p) {
  read(p.first);
  read(p.second);
}
template <>
void read(int& x) {
  scanf("%d", &x);
}
template <>
void read(unsigned int& x) {
  scanf("%u", &x);
}
template <>
void read(long long& x) {
  scanf("%lld", &x);
}
template <>
void read(unsigned long long& x) {
  scanf("%llu", &x);
}
template <typename T>
void readv(vector<T>& v) {
  int N;
  read(N);
  v.resize(N);
  read(v);
}
template <typename T, typename... Ts>
void read(T& x, Ts&... args) {
  read(x);
  read(args...);
}
template <typename T>
void imfix(T& r) {
  if (r >= 1000000007) r -= 1000000007;
}
template <typename T>
T mfix(T r) {
  return r >= 1000000007 ? r - 1000000007 : r;
}
int mpow(int a, int b) {
  long long r = 1;
  long long c = a;
  if (b == 0) return 1;
  while (b > 1) {
    if (b & 1) r = r * c % 1000000007;
    c = c * c % 1000000007;
    b >>= 1;
  }
  return r * c % 1000000007;
}
int inv(int a) { return mpow(a, 1000000007 - 2); }
const int half = inv(2);
int fact[200050];
int ifact[200050];
void fill_ccc() {
  fact[0] = 1;
  for (int i = 1; i < sizeof(fact) / sizeof(fact[0]); i++)
    fact[i] = (long long)fact[i - 1] * i % 1000000007;
  for (int i = 0; i < sizeof(fact) / sizeof(fact[0]); i++)
    ifact[i] = inv(fact[i]);
}
int cc(int n, int k) {
  return (long long)fact[n] * ifact[k] % 1000000007 * ifact[n - k] % 1000000007;
}
long long count(int mn, int mx, int n) {
  int ret = 0;
  for (int i = mn; i <= mx; i++) imfix(ret += cc(n, i));
  return (long long)mpow(half, n) * ret % 1000000007;
}
int main(int argc, const char* argv[]) {
  int N;
  long long T;
  fill_ccc();
  read(N, T);
  vector<long long> t(N);
  read(t);
  long long s = 0;
  long long ret = 0;
  for (int i = 0; i < N; s += t[i], i++) {
    if (s > T) break;
    for (int b = 0; b <= 1; b++) {
      long long tt = t[i] + b;
      long long mn = T - s - tt + 1;
      long long mx = T - s;
      mini<long long>(mx, i);
      maxi<long long>(mn, 0);
      if (mn > mx) continue;
      ret = (ret + i * count(mn, mx, i)) % 1000000007;
    }
  }
  if (s <= T) {
    long long mx = T - s;
    mini<long long>(mx, N);
    ret = (ret + 2 * N * count(0, mx, N)) % 1000000007;
  }
  printf("%lld\n", ret * half % 1000000007);
  return 0;
}
