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;
using ll = long long;
const ll INF = 1e15;
const ll MOD = 1e9 + 7;
int n;
ll T;
vector<ll> t;
vector<ll> fact;
ll mod_pow(ll b, ll e) {
  ll res = 1LL;
  while (e) {
    if (e & 1) res = (res * b) % MOD;
    b = (b * b) % MOD;
    e >>= 1;
  }
  return res;
}
ll mod_inv(ll den) { return mod_pow(den, MOD - 2); }
ll bin_coeff(ll n, ll k) {
  return (fact[n] * mod_inv(fact[k] * fact[n - k] % MOD)) % MOD;
}
int main() {
  scanf("%d %lld", &n, &T);
  t.resize(n + 2);
  fact.resize(n + 1);
  t[0] = 0;
  t[n + 1] = INF;
  for (int i = 1; i <= n; ++i) scanf("%lld", &t[i]);
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = (fact[i - 1] * i) % MOD;
  ll den = 1;
  ll sum = 0;
  ll ans = 0;
  for (int i = 0; i <= n; ++i) {
    den = (den << 1) % MOD;
    sum += t[i];
    if (sum > T) break;
    ll thresh = T - t[i + 1];
    if (thresh >= sum && thresh <= sum + i)
      ans = (ans + (i * bin_coeff(i, thresh - sum)) % MOD * mod_inv(den)) % MOD;
    for (ll j = max(sum, thresh + 1); j <= min(sum + i, T); ++j)
      ans = (ans + (2 * i * bin_coeff(i, j - sum)) % MOD * mod_inv(den)) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
