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 M = 2e5 + 5, JYY = 1e9 + 7;
long long max(long long x, long long y) { return x > y ? x : y; }
long long min(long long x, long long y) { return x < y ? x : y; }
long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') y = ch == '-' ? -1 : 1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * y;
}
long long fac[M], ifac[M], t[M], pw2[M];
long long Comb(long long a, long long b) {
  return fac[a] * ifac[b] % JYY * ifac[a - b] % JYY;
}
void solve() {
  long long n = read(), T = read(), Ans = 0;
  fac[0] = fac[1] = ifac[0] = ifac[1] = pw2[0] = 1, pw2[1] = (JYY + 1) / 2;
  for (long long i = 2; i <= n; i++)
    ifac[i] = (JYY - JYY / i) * ifac[JYY % i] % JYY,
    pw2[i] = pw2[i - 1] * pw2[1] % JYY;
  for (long long i = 2; i <= n; i++)
    fac[i] = fac[i - 1] * i % JYY, ifac[i] = ifac[i - 1] * ifac[i] % JYY;
  for (long long i = 1; i <= n; i++) t[i] = t[i - 1] + read();
  for (long long i = 1, res = 1, j = 0; i <= n; i++) {
    if (t[i] > T) break;
    long long r = min(T - t[i], i);
    res = (res + res - Comb(i - 1, j) + JYY) % JYY;
    while (j < r) res = (res + Comb(i, ++j)) % JYY;
    while (j > r) res = (res - Comb(i, j--) + JYY) % JYY;
    Ans = (Ans + res * pw2[i]) % JYY;
  }
  printf("%lld\n", Ans);
}
signed main() { solve(); }
