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 N = 2e5 + 20;
const int oo = 1e9 + 20;
const int MOD = 1e9 + 7;
inline int mul(int a, int b) { return 1ll * a * b % MOD; }
inline int add(int a, int b) { return a + b < MOD ? a + b : a + b - MOD; }
int fpow(int b, int e) {
  int ans = 1;
  for (; e; e >>= 1, b = mul(b, b))
    if (e & 1) ans = mul(ans, b);
  return ans;
}
int fact[N], ifact[N];
int comb(int n, int k) { return mul(fact[n], mul(ifact[k], ifact[n - k])); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  fact[0] = ifact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(i, fact[i - 1]);
    ifact[i] = fpow(fact[i], MOD - 2);
  }
  int n;
  long long T;
  while (cin >> n >> T) {
    vector<long long> acc(n);
    for (int i = 0; i < n; i++) {
      cin >> acc[i];
      if (i) acc[i] += acc[i - 1];
    }
    acc.push_back(1e18);
    int ans = 0, pot2 = 2;
    for (int i = 0; i < n; i++) {
      int cur = 0;
      if (i < n - 1) {
        pot2 = mul(pot2, 2);
        long long to = min(i + 1ll, T - acc[i]);
        long long from = max(0ll, T - acc[i + 1] + 1);
        for (long long k = from; k <= to; k++) cur = add(cur, comb(i + 1, k));
        to = min(i + 1ll, T - acc[i]);
        from = max(0ll, T - acc[i + 1]);
        for (long long k = from; k <= to; k++) cur = add(cur, comb(i + 1, k));
      } else {
        long long to = min(i + 1ll, T - acc[i]);
        long long from = max(0ll, T - acc[i + 1] + 1);
        for (long long k = from; k <= to; k++) cur = add(cur, comb(i + 1, k));
      }
      cur = mul(cur, fpow(pot2, MOD - 2));
      cur = mul(cur, i + 1);
      ans = add(cur, ans);
    }
    cout << ans << '\n';
  }
}
