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 mod = 1000000007;
const long long INF = mod * mod;
const long double eps = 1e-8;
long long mod_pow(long long x, long long n) {
  long long ret = 1;
  while (n > 0) {
    if (n % 2) ret = ret * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
struct perm {
 private:
  int sz;
  vector<long long> p, invp;

 public:
  perm(int n) {
    sz = n + 1;
    p.resize(sz), invp.resize(sz);
    p[0] = 1;
    for (int i = 1; i <= sz - 1; i++) {
      p[i] = p[i - 1] * i % mod;
    }
    invp[sz - 1] = 1;
    long long cop = mod - 2, x = p[sz - 1];
    while (cop) {
      if (cop % 2) invp[sz - 1] = invp[sz - 1] * x % mod;
      cop >>= 1;
      x = x * x % mod;
    }
    for (int i = sz - 1 - 1; i >= 0; i--) {
      invp[i] = invp[i + 1] * (i + 1) % mod;
    }
  }
  long long comb(long long x, long long y) {
    if (x < y || y < 0) return 0;
    long long ret = p[x];
    (ret *= invp[y]) %= mod;
    (ret *= invp[x - y]) %= mod;
    return ret;
  }
  long long combP(long long x, long long y) {
    if (x < y || y < 0) return 0;
    return p[x] * invp[x - y] % mod;
  }
};
perm p(1 << 19);
void solve() {
  int n;
  long long t;
  cin >> n >> t;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  a.push_back(INF);
  n++;
  vector<long long> ra(n + 1);
  for (int i = 0; i < n; i++) ra[i + 1] = ra[i] + a[i];
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    long long ri = t - ra[i];
    if (ri > i) ri = i;
    long long le = t - ra[i] - (a[i] - 1);
    if (le < 0) le = 0;
    long long csum = 0;
    for (long long j = le; j <= ri; j++) {
      csum += p.comb(i, j);
      if (csum >= mod) csum -= mod;
    }
    long long b = mod_pow(2, i);
    b = mod_pow(b, mod - 2);
    csum = csum * b % mod;
    ans += csum * i % mod;
    if (ans >= mod) ans -= mod;
    long long dif = t - ra[i + 1];
    if (0 <= dif && dif <= i) {
      long long ad = p.comb(i, dif) * b % mod * (1 + mod) / 2 % mod;
      ans += ad * i % mod;
      if (ans >= mod) ans -= mod;
    }
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
