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("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const long long N = 2e5 + 5;
const long long M = 70;
const long long SZ = 450;
const long long mod = 1e9 + 7;
const long long inf = 0x3f3f3f3f3f3f3f3f;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
long long ksm(long long x, long long y = mod - 2, long long z = mod) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
long long inv[N], fac[N], ifc[N];
void Init(long long n) {
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  ifc[0] = 1;
  for (long long i = 1; i <= n; i++) ifc[i] = ifc[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  if (m < 0 || m > n) return 0;
  return fac[n] * ifc[m] % mod * ifc[n - m] % mod;
}
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
long long n, T, s, ans, cnt, res;
long long t[N];
long long x, y, z, vis;
signed main() {
  Init(N - 1);
  n = read();
  T = read();
  for (long long i = 1; i <= n; i++) t[i] = read();
  for (long long i = 1; i <= n; i++) {
    s += t[i];
    if (T - s >= i) {
      ans = (ans + ksm(2, n - i) * ksm(2, i)) % mod;
      continue;
    } else if (T - s < 0)
      break;
    if (!vis) x = i, y = i, z = ksm(2, i), vis = 1;
    while (x < i) z = (2 * z - C(x, y)) % mod, x++;
    while (y > T - s) z = (z - C(x, y)) % mod, y--;
    ans = (ans + z * ksm(2, n - i)) % mod;
  }
  cout << (ans * ksm(ksm(2, n)) % mod + mod) % mod << "\n";
}
