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 N = 2e5 + 10, mod = 1e9 + 7;
long long n, T, ans;
long long t[N], pre[N], p[N];
long long fac[N], inv[N];
long long qpow(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
  }
  return ret;
}
void prework(int n) {
  fac[0] = inv[0] = inv[1] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  for (long long i = 2; i <= n; i++)
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  for (long long i = 2; i <= n; i++) inv[i] = inv[i] * inv[i - 1] % mod;
}
long long C(long long n, long long m) {
  return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
int main() {
  cin >> n >> T;
  prework(n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &t[i]);
    pre[i] = pre[i - 1] + t[i];
  }
  t[++n] = T + 1, pre[n] = pre[n - 1] + t[n];
  for (long long i = 0; i <= n - 1; i++) {
    if (pre[i] > T) break;
    if (pre[i + 1] + i + 1 <= T) continue;
    long long pw = qpow(qpow(2, i + 1), mod - 2);
    long long l = max(0ll, T - pre[i + 1] + 1), r = min(T - pre[i], i);
    for (long long k = l; k <= r; k++)
      (ans += 1ll * i * C(i, k) % mod * pw % mod) %= mod;
    l = max(0ll, T - pre[i + 1]), r = min(T - pre[i], i);
    for (long long k = l; k <= r; k++)
      (ans += 1ll * i * C(i, k) % mod * pw % mod) %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
