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 MAXN = 300001;
long long t;
int n;
long long s[MAXN];
int fac[MAXN], revfac[MAXN], pow2[MAXN];
int mod = (int)(1e9) + 7;
int ksm(int x, int y) {
  int sum = 1;
  for (; y; y >>= 1) {
    if (y & 1) sum = 1LL * sum * x % mod;
    x = 1LL * x * x % mod;
  }
  return sum;
}
int C(int nn, int kk) {
  return 1LL * fac[nn] * revfac[kk] % mod * revfac[nn - kk] % mod;
}
int main() {
  scanf("%d%I64d", &n, &t);
  int bound = n;
  s[0] = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &s[i]);
    s[i] += s[i - 1];
    if (s[i] > t and bound == n) bound = i - 1;
  }
  fac[0] = revfac[0] = pow2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = 1LL * i * fac[i - 1] % mod;
    pow2[i] = 2LL * pow2[i - 1] % mod;
  }
  for (int i = 0; i <= n; ++i) {
    revfac[i] = ksm(fac[i], mod - 2);
    pow2[i] = ksm(pow2[i], mod - 2);
  }
  int delta = 0;
  int ans = 0;
  while (bound >= delta) {
    if (bound == n)
      delta = 0;
    else if (t - s[bound + 1] < 0)
      delta = 0;
    else {
      delta = t - s[bound + 1];
      if (delta > bound) break;
      ans += 1LL * pow2[bound + 1] * C(bound, delta) % mod * bound % mod;
      ans %= mod;
      ++delta;
    }
    while (delta <= t - s[bound] and delta <= bound) {
      ans += 1LL * pow2[bound] * C(bound, delta) % mod * bound % mod;
      ans %= mod;
      ++delta;
    }
    --bound;
    delta = 0;
  }
  printf("%d", ans);
  return 0;
}
