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 = 2e5 + 10;
long long fac[maxn], ifac[maxn], p[maxn], ip[maxn];
long long t[maxn];
int N;
long long T;
long long qpow(long long x, long long n) {
  if (n <= 0) return 1;
  long long ret = qpow(x, n >> 1);
  ret = ret * ret % mod;
  if (n & 1) ret = ret * x % mod;
  return ret;
}
long long C(long long n, long long r) {
  if (r > n || n < 0 || r < 0) return 0;
  return fac[n] * (ifac[r] * ifac[n - r] % mod) % mod;
}
void init() {
  fac[0] = p[0] = 1;
  ifac[0] = qpow(fac[0], mod - 2);
  ip[0] = qpow(p[0], mod - 2);
  for (int i = 1; i < maxn; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    p[i] = p[i - 1] * 2 % mod;
    ifac[i] = qpow(fac[i], mod - 2);
    ip[i] = qpow(p[i], mod - 2);
  }
}
int main() {
  init();
  scanf("%d %lld", &N, &T);
  for (int i = 1; i <= N; ++i) {
    scanf("%lld", &t[i]);
  }
  long long sum = 0, ret = 0, prefix = 0;
  int pre = -1;
  for (int i = 1; i <= N; ++i) {
    sum += t[i];
    if (sum > T) break;
    if (sum + i <= T) {
      ret = (ret + 1) % mod;
      continue;
    }
    int K = min((long long)i, T - sum);
    if (prefix == 0) {
      for (int j = 0; j <= K; ++j) {
        prefix = (prefix + C(i, j)) % mod;
      }
      pre = K;
      ret = (ret + prefix * ip[i] % mod) % mod;
    } else {
      prefix = prefix * 2 % mod;
      prefix = (prefix - C(i - 1, pre) + mod) % mod;
      while (pre > K) {
        prefix = (prefix - C(i, pre) + mod) % mod;
        --pre;
      }
      ret = (ret + prefix * ip[i] % mod) % mod;
    }
  }
  printf("%lld\n", ret);
}
