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 int mod = 1000000007;
int add(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
int sub(int a, int b) { return a < b ? a - b + mod : a - b; }
int mul(int a, int b) { return 1LL * a * b % mod; }
int mul(int a, int b, int c) { return mul(mul(a, b), c); }
const int maxn = 1 << 18;
int inv[maxn], fact[maxn], finv[maxn], inv2[maxn];
void init(int n = maxn) {
  inv[0] = inv[1] = fact[0] = fact[1] = finv[0] = finv[1] = inv2[0] = 1;
  inv2[1] = (mod + 1) >> 1;
  for (int i = 2; i != n; ++i) {
    inv[i] = mul(inv[mod % i], mod - mod / i);
    fact[i] = mul(fact[i - 1], i);
    finv[i] = mul(finv[i - 1], inv[i]);
    inv2[i] = mul(inv2[i - 1], inv[2]);
  }
}
int bino(int n, int k) { return mul(fact[n], finv[k], finv[n - k]); }
int last_n, last_k, last_c;
int calc(int n, int64_t k) {
  if (!last_n) {
    last_n = n;
    last_k = k;
    for (int i = 0; i <= k; ++i) last_c = add(last_c, bino(n, i));
    return last_c;
  }
  while (last_k > k) last_c = sub(last_c, bino(last_n, last_k--));
  while (last_n < n) last_c = sub(add(last_c, last_c), bino(last_n++, last_k));
  return last_c;
}
int64_t t[maxn];
int main() {
  init();
  int n;
  int64_t T;
  cin >> n >> T;
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    t[i] += t[i - 1];
    if (t[i] <= T)
      res = add(res, T - t[i] > i ? 1 : mul(calc(i, T - t[i]), inv2[i]));
  }
  cout << res << endl;
}
