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;
namespace IO {
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
}  // namespace IO
namespace Utility {
template <typename T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <typename T>
T lcm(T a, T b) {
  return (a / gcd(a, b) * b);
}
template <typename T>
inline T addmod(T a, T b, T mod) {
  T c = a + b;
  return c > mod ? c - mod : c;
}
template <typename T>
inline T submod(T a, T b, T mod) {
  T c = a - b;
  return c < mod ? c + mod : c;
}
template <typename T>
inline T mulmod(T a, T b, T mod) {
  return a * 1LL * b % mod;
}
template <typename T>
T exEuclid(T a, T b, T &x, T &y) {
  x = 0, y = 1;
  if (a == 0) return b;
  T x1, y1;
  T d = exEuclid(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
template <typename T>
T modInv(T a, T mod) {
  T x, y;
  exEuclid(a, mod, x, y);
  return x < 0 ? x + mod : x;
}
template <typename T>
T bigmod(T b, T p, T mod) {
  T ret = b > 0;
  while (p) {
    if (p & 1) ret = mulmod(ret, b, mod);
    p >>= 1, b = mulmod(b, b, mod);
  }
  return ret;
}
}  // namespace Utility
using namespace IO;
using namespace Utility;
const int INF = 2000000009;
const int MX = 200005;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000000007;
int fac[MX], ifac[MX], inv[MX], pw[MX];
int ncr(int n, int r) {
  if (r > n || r < 0) return 0;
  return (int)(1LL * fac[n] * ifac[r] % MOD * ifac[n - r] % MOD);
}
int main() {
  fac[0] = fac[1] = ifac[0] = ifac[1] = inv[0] = inv[1] = 1;
  pw[0] = 1, pw[1] = 2;
  for (int i = 2; i < MX; i++) {
    fac[i] = 1LL * fac[i - 1] * i % MOD;
    inv[i] = MOD - 1LL * inv[MOD % i] * (MOD / i) % MOD;
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % MOD;
    pw[i] = pw[i - 1] * 2 % MOD;
  }
  int n = ({
    int a;
    read(a);
    a;
  });
  long long T = ({
    long long a;
    read(a);
    a;
  }),
            t = 0LL;
  int sum = 0LL;
  int k = 0, ways = 1;
  for (int i = 1; i <= n; i++) {
    t += ({
      long long a;
      read(a);
      a;
    });
    if (t > T) break;
    int cur_k = (int)min(1LL * i, T - t);
    while (k > cur_k) {
      ways = (ways - ncr(i - 1, k--));
      if (ways < 0) ways += MOD;
    }
    while (k < cur_k) {
      ways = (ways + ncr(i - 1, ++k)) % MOD;
    }
    ways = ways * 2 % MOD;
    ways -= ncr(i - 1, k);
    if (ways < 0) ways += MOD;
    sum = (sum + 1LL * ways * pw[n - i] % MOD) % MOD;
  }
  int ans = 1LL * sum * modInv(pw[n], MOD) % MOD;
  printf("%d\n", ans);
  return 0;
}
