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 int T = 1e9 + 7;
const int maxn = 2e5 + 233;
long long fac[maxn], inv[maxn];
int exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return d;
}
inline void add(long long &x, long long v) { x = (x + v) % T; }
inline int Inv(int i) {
  int x, y;
  exgcd(i, T, x, y);
  x = (x + T) % T;
  return x;
}
void init() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i < maxn; i++) {
    fac[i] = fac[i - 1] * i % T;
    int x, y;
    exgcd(i, T, x, y);
    x = (x + T) % T;
    inv[i] = inv[i - 1] * x % T;
  }
}
long long inv2[maxn];
inline long long c(int n, int k) {
  if (k > n) return 0;
  return fac[n] * inv[k] % T * inv[n - k] % T * inv2[n] % T;
}
int t[maxn];
int main() {
  init();
  int n;
  long long tot;
  cin >> n >> tot;
  for (int i = 1; i <= n; i++) scanf("%d", &t[i]);
  long long cur = 0, ans = 0;
  inv2[1] = (T + 1) / 2;
  for (int i = 2; i <= n; i++) inv2[i] = (inv2[i - 1] * (T + 1) / 2) % T;
  for (int i = 1; i <= n; i++) {
    cur += t[i];
    if (cur > tot) break;
    if (cur + t[i + 1] + i < tot) continue;
    int l = tot - cur - t[i + 1], r = tot - cur;
    if (r > i) r = i;
    if (l >= 0) {
      if (i != n) {
        add(ans, c(i, l) * i % T * inv2[1] % T);
        l++;
      }
    } else
      l = 0;
    if (i == n) l = 0;
    for (int j = l; j <= r; j++) add(ans, c(i, j) * i % T);
  }
  cout << ans << endl;
  return 0;
}
