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;
bool cmp(int a, int b) { return a > b; }
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1);
const int N = 2e5 + 10;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % mod;
  if (a < 0) a += mod;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % mod;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <typename T>
inline int bcnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
inline long long fpow(long long a, long long b) {
  long long res = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long last = 1, ln, lk;
long long a[N];
long long fac[N];
long long inv[N];
long long C(int n, int m) { return fac[n] * inv[m] % mod * inv[n - m] % mod; }
void init() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i < N; i++)
    fac[i] = fac[i - 1] * i % mod, inv[i] = fpow(fac[i], mod - 2);
}
long long fun(long long n, long long k) {
  while (ln < n) Adm(last, last), Adm(last, -C(ln, lk)), ln++, lk++;
  while (lk > k) lk--, Adm(last, C(ln, lk));
  while (lk < k) Adm(last, -C(ln, lk)), lk++;
  return last;
}
int main() {
  init();
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]), a[i] += a[i - 1];
  long long ans = 0, p = 1;
  for (int i = 1; i <= n; i++) {
    Mum(p, inv[2]);
    if (a[i] > m)
      break;
    else if (a[i] + i <= m)
      Adm(ans, 1);
    else
      Adm(ans, fun(i, i + a[i] - m) * p);
  }
  cout << (ans + mod) % mod << endl;
  return 0;
}
