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>
#pragma GCC optimize(2)
using namespace std;
const int maxn = 2e5;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double Pi = acos(-1.0);
inline int read_int() {
  char c;
  int ret = 0, sgn = 1;
  do {
    c = getchar();
  } while ((c < '0' || c > '9') && c != '-');
  if (c == '-')
    sgn = -1;
  else
    ret = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  return sgn * ret;
}
inline long long read_ll() {
  char c;
  long long ret = 0, sgn = 1;
  do {
    c = getchar();
  } while ((c < '0' || c > '9') && c != '-');
  if (c == '-')
    sgn = -1;
  else
    ret = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  return sgn * ret;
}
inline long long quick_pow(long long base, long long index, long long p) {
  long long res = 1;
  while (index) {
    if (index & 1) res = res * base % p;
    base = base * base % p;
    index >>= 1;
  }
  return res;
}
long long fac[maxn + 5], inv[maxn + 5], inv2[maxn + 5], t[maxn + 5], T, sum;
long long C(long long a, long long b) {
  return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
long long res, ans;
int n;
int main() {
  n = read_int(), T = read_ll();
  for (int i = 1; i <= n; ++i) t[i] = read_ll();
  fac[0] = inv[0] = inv[1] = inv2[0] = 1;
  for (int i = 1; i <= maxn; ++i) fac[i] = fac[i - 1] * i % mod;
  for (int i = 2; i <= maxn; ++i) inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  for (int i = 1; i <= maxn; ++i) inv2[i] = inv2[i - 1] * inv[2] % mod;
  for (int i = 1; i <= maxn; ++i) inv[i] = inv[i - 1] * inv[i] % mod;
  long long pre = 0;
  ans = 1;
  for (int i = 1; i <= n; ++i) {
    t[i] += t[i - 1];
    if (t[i] <= T) {
      if (T - t[i] > i)
        res = (res + 1) % mod;
      else {
        long long u = T - t[i];
        while (pre < u) {
          ++pre;
          ans = (ans + C(i, pre)) % mod;
        }
        while (pre > u) {
          ans = (ans + mod - C(i, pre)) % mod;
          --pre;
        }
        res = (res + ans * inv2[i] % mod) % mod;
      }
    }
    ans = (2 * ans + mod - C(i, pre)) % mod;
  }
  printf("%lld\n", res);
  return 0;
}
