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 twx = 2e5 + 100;
const int MOD = 1000000007;
long long l, r;
long long S;
long long t[twx];
long long c[twx];
long long fc[twx];
long long n, m;
long long z;
long long ans;
long long asd1, asd2;
long long i, j;
long long read() {
  long long sum = 0;
  long long flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') {
      flag = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = ((sum * 10) + c - '0');
    c = getchar();
  }
  return sum * flag;
}
void PLUS(long long &a, long long b) {
  a += b;
  a -= a > MOD ? MOD : 0;
}
int POW(int z, int k) {
  int qwe = 1;
  for (; k; z = (long long)z * z % MOD, k >>= 1) {
    if (k & 1) {
      qwe = (long long)qwe * z % MOD;
    }
  }
  return qwe;
}
long long C(long long n, long long m) {
  if (m < 0 || m > n) {
    return 0;
  }
  return (long long)c[n] * fc[m] % MOD * fc[n - m] % MOD;
}
void init() {
  n = read();
  m = read();
  for (i = 1; i <= n; i++) {
    t[i] = read();
  }
  for (c[0] = i = 1; i <= n; i++) {
    c[i] = (long long)c[i - 1] * i % MOD;
  }
  for (fc[n] = POW(c[n], MOD - 2), i = n; i; --i) {
    fc[i - 1] = (long long)fc[i] * i % MOD;
  }
  asd2 = POW(2, MOD - 2);
  asd1 = 1;
  for (i = 1; i <= n; i++) {
    z = 0;
    asd1 = (long long)asd1 * asd2 % MOD;
    S += t[i];
    r = min(m - S, (long long)i);
    l = m - S - t[i + 1];
    if (i == n) {
      l = -1;
    }
    z = 0;
    for (j = max(l + 1, (long long)0); j <= r; j++) {
      PLUS(z, C(i, j));
    }
    if (l >= 0) {
      PLUS(z, (long long)C(i, l) * asd2 % MOD);
    }
    z = (long long)z * asd1 % MOD * i % MOD;
    PLUS(ans, z);
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  return 0;
}
