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 c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<char*> tokenizer(const char* args) {
  char* token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char*> v({token});
  while (token = strtok(NULL, ", ")) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
int64_t powm(int64_t a, int64_t b, int64_t mod) {
  int64_t res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long double pi = 3.14159265358979323846;
const int64_t mod = 1e9 + 7;
const int64_t N = 1000005;
const int64_t inf = 1e9;
const int64_t INF = 1e18;
void solve() {
  int64_t n, t;
  cin >> n >> t;
  vector<int64_t> a(n);
  for (int64_t fukre = 0; fukre < (n); fukre++) {
    cin >> a[fukre];
  };
  vector<int64_t> prefix(n);
  for (int64_t i = 0; i < n; i++) prefix[i] = a[i];
  for (int64_t i = 1; i < n; i++) prefix[i] = (prefix[i - 1] + prefix[i]);
  int64_t ans = 0;
  vector<int64_t> fact(n + 2, 1), invFact(n + 2);
  for (int64_t i = 1; i < n + 2; i++) fact[i] = fact[i - 1] * i % mod;
  invFact[n + 1] = powm(fact[n + 1], mod - 2, mod);
  for (int64_t i = n; i >= 0; i--) invFact[i] = invFact[i + 1] * (i + 1) % mod;
  auto nCr = [&](int64_t n, int64_t r) -> int64_t {
    if (r > n) return 0;
    return fact[n] * invFact[n - r] % mod * invFact[r] % mod;
  };
  int64_t sum = 0, ttt = 0, fl = 0;
  for (int64_t i = 0; i < n; i++) {
    if (prefix[i] + i + 1 <= t)
      ans = (ans + 1) % mod;
    else if (prefix[i] <= t) {
      int64_t k = t - prefix[i];
      k = min(k, i + 1);
      {};
      if (!fl) {
        for (int64_t j = 0; j <= k; j++) {
          sum = (sum + nCr(i + 1, j)) % mod;
        }
      } else {
        sum = (sum * 2) % mod;
        sum = (sum - nCr(i, ttt) + mod) % mod;
        for (int64_t j = ttt; j > k; j--)
          sum = (sum - nCr(i + 1, j) + mod) % mod;
      }
      ans = (ans + sum * powm(powm(2, i + 1, mod), mod - 2, mod) % mod) % mod;
      ttt = k;
      fl = 1;
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int64_t tc = 1;
  for (int64_t t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}
