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 long long mod = 1e9 + 7;
const int maxN = 200010;
long long a[maxN];
long long times = 1;
long long inv2;
long long presum[maxN];
int n;
long long T;
long long ans = 0;
long long powmod(long long base, long long times) {
  long long ret = 1;
  while (times) {
    if (times % 2) ret = (ret * base) % mod;
    base = (base * base) % mod;
    times /= 2;
  }
  return ret % mod;
}
void init();
long long C(long long n, long long x) {
  long long ret = powmod(presum[x], mod - 2);
  ret = (ret * presum[n]) % mod;
  ret = (ret * powmod(presum[n - x], mod - 2)) % mod;
  return ret;
}
int main() {
  init();
  scanf("%d %lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  a[n + 1] = 3e14;
  for (long long i = 0; i <= n; i++) {
    T -= a[i];
    if (T < 0) break;
    times = (inv2 * times) % mod;
    long long tmp = T - a[i + 1] + 1;
    if (tmp > i + 1) continue;
    long long cur = 0;
    for (long long j = max(0ll, tmp); j <= min(i, T); j++) cur += C(i, j);
    for (long long j = max(0ll, tmp - 1); j <= min(i, T); j++) cur += C(i, j);
    cur %= mod;
    cur = (cur * times) % mod;
    cur = (cur * i) % mod;
    ans = (ans + cur) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
void init() {
  inv2 = powmod(2, mod - 2);
  presum[0] = 1;
  for (long long i = 1; i <= 200000; i++) presum[i] = (presum[i - 1] * i) % mod;
}
