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 MAXN = 200015;
const long long mod = 1e9 + 7;
template <typename T>
void inc(T &x, T y) {
  x = x + y >= mod ? x + y - mod : x + y;
}
template <typename T>
void read(T &x) {
  T flag = 1;
  char ch = getchar();
  for (; '0' > ch || ch > '9'; ch = getchar())
    if (ch == '-') flag = -1;
  for (x = 0; '0' <= ch && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  x *= flag;
}
void file() {}
long long n, T;
long long t[MAXN], st[MAXN], Pow[MAXN], inv[MAXN], fac[MAXN], ifac[MAXN];
long long C(long long x, long long y) {
  return x < 0 || y < 0 || x < y ? 0
                                 : fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
map<long long, long long> s[MAXN];
long long S(long long x, long long y) {
  if (s[x].find(y) != s[x].end()) return s[x][y];
  long long ret = 0;
  if (x <= y)
    ret = S(x, x);
  else
    ret = ((S(x, y + 1) - C(x, y + 1)) % mod + mod) % mod;
  return s[x][y] = ret;
}
int main() {
  file();
  read(n);
  read(T);
  inv[0] = inv[1] = fac[0] = ifac[0] = 1;
  for (long long i = 2; i <= n + 10; i++)
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  for (long long i = 1; i <= n + 10; i++)
    fac[i] = fac[i - 1] * i % mod, ifac[i] = ifac[i - 1] * inv[i] % mod;
  Pow[0] = 1;
  for (long long i = 1; i <= n + 10; i++) Pow[i] = Pow[i - 1] * inv[2] % mod;
  for (long long i = 1; i <= n; i++) read(t[i]), st[i] = st[i - 1] + t[i];
  if (st[1] > T) {
    puts("0");
    return 0;
  }
  s[1][0] = 1;
  s[1][1] = 2;
  long long ans = s[1][min(1ll, T - st[1])] * Pow[1] % mod;
  for (long long i = 2; i <= n && st[i] <= T; i++) {
    if (st[i] <= T) {
      long long m = min(T - st[i], (long long)i);
      s[i][m] = ((2 * S(i - 1, m) - C(i - 1, m)) % mod + mod) % mod;
      inc(ans, s[i][m] * Pow[i] % mod);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
