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 = 300005, maxm = 515;
const double pi = acos(-1.0);
const long long mod = 1000000007;
long long ksm(long long a, long long k) {
  long long r = 1;
  while (k > 0) {
    if (k & 1) r = r * a % mod;
    a = a * a % mod;
    k >>= 1;
  }
  return r;
}
int n;
long long t;
int a[maxn];
long long jci[maxn] = {1, 0}, jc[maxn] = {1, 0}, iv2;
int main() {
  int i, j, k;
  iv2 = ksm(2, mod - 2);
  for (i = 1; i < maxn; i++) {
    jc[i] = jc[i - 1] * i % mod;
    jci[i] = ksm(jc[i], mod - 2);
  }
  scanf("%d %I64d", &n, &t);
  for (i = 0; i < n; i++) scanf("%d", a + i);
  long long ans = 0, ct = t, ctt = t, bw;
  for (i = 0; i < n; i++) {
    ct -= a[i];
    if (ct - i - 1 >= 0) {
      ans++;
      ctt = ct;
      continue;
    }
    if (ct < 0) break;
    if (ctt >= i) {
      bw = 0;
      for (j = 0; j <= ct; j++)
        bw += jc[i + 1] * jci[j] % mod * jci[i + 1 - j] % mod, bw %= mod;
    } else {
      bw = bw * 2 % mod + mod - jc[i] * jci[ctt] % mod * jci[i - ctt] % mod;
      bw %= mod;
      for (j = ctt; j > ct; j--) {
        bw += mod - jc[i + 1] * jci[j] % mod * jci[i + 1 - j] % mod;
        bw %= mod;
      }
    }
    ans += bw * ksm(iv2, i + 1) % mod;
    ans %= mod;
    ctt = ct;
  }
  printf("%I64d\n", ans);
  return 0;
}
