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;
namespace FGF {
int n;
long long m;
const int N = 2e5 + 5, mo = 1e9 + 7, inv2 = (mo + 1) / 2;
long long a[N];
int fac[N], inv[N], p[N], ans;
long long C(int x, int y) {
  if (x < 0 || y < 0 || x < y) return 0;
  return 1ll * fac[x] * inv[y] % mo * inv[x - y] % mo;
}
void work() {
  scanf("%d%lld", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), a[i] += a[i - 1];
  inv[0] = inv[1] = fac[0] = 1;
  for (int i = 2; i <= n; i++) inv[i] = 1ll * inv[mo % i] * (mo - mo / i) % mo;
  for (int i = 1; i <= n; i++)
    inv[i] = 1ll * inv[i] * inv[i - 1] % mo, fac[i] = 1ll * fac[i - 1] * i % mo;
  for (int i = 1, cur = 0, sum = 1, i2 = inv2; i <= n;
       i++, i2 = 1ll * inv2 * i2 % mo) {
    int lim = min((long long)i, m - a[i]);
    if (lim < 0) break;
    sum = (2ll * sum % mo - C(i - 1, cur) + mo) % mo;
    while (cur < lim) cur++, sum = (sum + C(i, cur)) % mo;
    while (cur > lim) sum = (sum - C(i, cur) + mo) % mo, cur--;
    p[i] = 1ll * i2 * sum % mo;
  }
  for (int i = 0; i <= n; i++)
    ans = (ans + 1ll * i * (p[i] - p[i + 1] + mo) % mo) % mo;
  printf("%d\n", ans);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
