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;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 2e6 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
ll t[MAXN], fac[MAXN], rfac[MAXN];
ll r2[MAXN];
ll FastPow(ll x, ll exp) {
  ll ret = 1;
  while (exp) {
    if (exp & 1) ret = ret * x % MOD;
    x = x * x % MOD;
    exp >>= 1;
  }
  return ret;
}
void Init() {
  fac[0] = fac[1] = 1;
  for (int i = (int)(1); i < (int)(MAXN); ++i) {
    fac[i] = fac[i - 1] * i % MOD;
  }
  rfac[MAXN - 1] = FastPow(fac[MAXN - 1], MOD - 2);
  for (int i = (int)(MAXN - 2); i >= (int)(0); --i) {
    rfac[i] = rfac[i + 1] * (i + 1) % MOD;
  }
  r2[0] = 1;
  r2[1] = FastPow(2, MOD - 2);
  for (int i = (int)(2); i < (int)(MAXN); ++i) {
    r2[i] = r2[i - 1] * r2[1] % MOD;
  }
}
inline ll C(ll n, ll x) { return fac[n] * rfac[x] % MOD * rfac[n - x] % MOD; }
ll Solution() {
  Init();
  ll tot = t[0], cnt = 1;
  ll ans = 0;
  while (tot <= T && cnt <= N) {
    if (cnt < N && tot + t[cnt] + cnt + 1 <= T) {
    } else {
      if (cnt == N) {
        ll rem = T - tot;
        for (int i = (int)(0); i < (int)(rem + 1); ++i) {
          ans += cnt * C(cnt, i) % MOD * r2[cnt] % MOD;
          ans %= MOD;
        }
      } else {
        ll g = tot + t[cnt];
        ll rem = T - g;
        ll lb = max(0LL, rem + 1), rb = min(cnt, T - tot);
        for (int i = (int)(lb); i < (int)(rb + 1); ++i) {
          ans += cnt * C(cnt, i) % MOD * r2[cnt + 1] % MOD;
          ans %= MOD;
        }
        rem = T - (g + 1);
        lb = max(0LL, rem + 1), rb = min(cnt, T - tot);
        for (int i = (int)(lb); i < (int)(rb + 1); ++i) {
          ans += cnt * C(cnt, i) % MOD * r2[cnt + 1] % MOD;
          ans %= MOD;
        }
      }
    }
    tot += t[cnt++];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N >> T) {
    for (int i = (int)(0); i < (int)(N); ++i) cin >> t[i];
    cout << Solution() << '\n';
  }
  return 0;
}
