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;
;
int MOD = 1e9 + 7;
struct mod_int {
  int val;
  mod_int(long long v = 0) {
    if (v < 0) v = v % MOD + MOD;
    if (v >= MOD) v %= MOD;
    val = v;
  }
  static int mod_inv(int a, int m = MOD) {
    int g = m, r = a, x = 0, y = 1;
    while (r != 0) {
      int q = g / r;
      g %= r;
      swap(g, r);
      x -= q * y;
      swap(x, y);
    }
    return x < 0 ? x + m : x;
  }
  explicit operator int() const { return val; }
  mod_int &operator+=(const mod_int &other) {
    val += other.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  mod_int &operator-=(const mod_int &other) {
    val -= other.val;
    if (val < 0) val += MOD;
    return *this;
  }
  static unsigned fast_mod(uint64_t x, unsigned m = MOD) {
    return x % m;
    unsigned x_high = x >> 32, x_low = (unsigned)x;
    unsigned quot, rem;
    asm("divl %4\n" : "=a"(quot), "=d"(rem) : "d"(x_high), "a"(x_low), "r"(m));
    return rem;
  }
  mod_int &operator*=(const mod_int &other) {
    val = fast_mod((uint64_t)val * other.val);
    return *this;
  }
  mod_int &operator/=(const mod_int &other) { return *this *= other.inv(); }
  friend mod_int operator+(const mod_int &a, const mod_int &b) {
    return mod_int(a) += b;
  }
  friend mod_int operator-(const mod_int &a, const mod_int &b) {
    return mod_int(a) -= b;
  }
  friend mod_int operator*(const mod_int &a, const mod_int &b) {
    return mod_int(a) *= b;
  }
  friend mod_int operator/(const mod_int &a, const mod_int &b) {
    return mod_int(a) /= b;
  }
  mod_int &operator++() {
    val = val == MOD - 1 ? 0 : val + 1;
    return *this;
  }
  mod_int &operator--() {
    val = val == 0 ? MOD - 1 : val - 1;
    return *this;
  }
  mod_int operator++(int) {
    mod_int before = *this;
    ++*this;
    return before;
  }
  mod_int operator--(int) {
    mod_int before = *this;
    --*this;
    return before;
  }
  mod_int operator-() const { return val == 0 ? 0 : MOD - val; }
  bool operator==(const mod_int &other) const { return val == other.val; }
  bool operator!=(const mod_int &other) const { return val != other.val; }
  mod_int inv() const { return mod_inv(val); }
  mod_int pow(long long p) const {
    assert(p >= 0);
    mod_int a = *this, result = 1;
    while (p > 0) {
      if (p & 1) result *= a;
      a *= a;
      p >>= 1;
    }
    return result;
  }
  friend ostream &operator<<(ostream &stream, const mod_int &m) {
    return stream << m.val;
  }
  friend istream &operator>>(istream &stream, mod_int &m) {
    return stream >> m.val;
  }
};
const int N = 2e5 + 10;
mod_int p[N], fac[N], inv[N];
map<pair<int, int>, mod_int> dp;
int pre;
mod_int ncr(int n, int r) { return fac[n] * inv[r] * inv[n - r]; }
mod_int tot_ncr(int n, int k) {
  if ((int)((dp).size()) == 0) {
    for (int i = 0; i <= k; i++) {
      dp[{n, k}] += ncr(n, i);
    }
    pre = k;
  } else {
    dp[{n, k}] = mod_int(2) * mod_int(dp[{n - 1, pre}]);
    dp[{n, k}] -= ncr(n - 1, pre);
    while (pre > k) {
      dp[{n, k}] -= ncr(n, pre);
      pre--;
    }
  }
  return dp[{n, k}];
}
void solve() {
  int n;
  long long T;
  mod_int ans = 2;
  cin >> n >> T;
  long long t[n + 1];
  t[0] = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    t[i] = t[i - 1] + t[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (t[i] + i <= T)
      p[i] = 1;
    else if (t[i] > T)
      p[i] = 0;
    else
      p[i] = tot_ncr(i, T - t[i]) / ans.pow(i);
  }
  cout << accumulate(p + 1, p + n + 1, mod_int(0)) << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(15);
  fac[0] = inv[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = fac[i - 1] * i;
    inv[i] = inv[i - 1] / i;
  }
  while (t--) {
    solve();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}
