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 long long mod = 1e9 + 7LL;
long long multiply(long long a, long long b, long long MOD) {
  long long res;
  res = (a % MOD) * (b % MOD);
  res %= MOD;
  return res;
}
long long add(long long a, long long b, long long MOD) {
  long long res = (a % MOD) + (b % MOD);
  res %= MOD;
  if (res < 0LL) res += MOD;
  return res;
}
long long power(long long base, long long exp, long long MOD) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = multiply(res, base, MOD);
    base = multiply(base, base, MOD);
    exp /= 2;
  }
  return res % MOD;
}
long long inv(long long n) { return power(n, mod - 2LL, mod); }
long long ncr(const vector<long long> &fact, const vector<long long> &iFact,
              long long n, long long r) {
  if (n == r || r == 0) return 1;
  return multiply(fact[n], multiply(iFact[r], iFact[n - r], mod), mod);
}
void solve() {
  long long n, T;
  cin >> n >> T;
  const int N = 200 * 1000 + 100;
  vector<long long> fact(N), iFact(N);
  fact[0] = 1;
  iFact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = multiply(fact[i - 1], i, mod);
    iFact[i] = multiply(iFact[i - 1], inv(i), mod);
  }
  vector<long long> t(n);
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  vector<long long> E(n, -1), pref(n);
  pref[0] = t[0];
  for (int i = 1; i < n; i++) {
    pref[i] = pref[i - 1] + t[i];
  }
  long long lastMi = -1;
  long long curPrefSum = -1;
  for (int i = 0; i < n; i++) {
    if (pref[i] > T) {
      E[i] = 0;
    } else if (pref[i] + i + 1LL <= T) {
      E[i] = 1;
    } else {
      if (curPrefSum == -1) {
        lastMi = min(T - pref[i], i + 1LL);
        curPrefSum = 0;
        for (int j = 0; j <= lastMi; j++) {
          curPrefSum = add(curPrefSum, ncr(fact, iFact, i + 1LL, j), mod);
        }
        E[i] = multiply(curPrefSum, inv(power(2LL, i + 1, mod)), mod);
      } else {
        lastMi++;
        curPrefSum = multiply(curPrefSum, 2LL, mod);
        curPrefSum = add(curPrefSum, ncr(fact, iFact, i, lastMi), mod);
        long long termsNeeded = min(T - pref[i], i + 1LL);
        if (termsNeeded < lastMi) {
          while (lastMi > termsNeeded) {
            long long term = ncr(fact, iFact, i + 1LL, lastMi);
            curPrefSum -= term;
            curPrefSum %= mod;
            if (curPrefSum < 0) curPrefSum += mod;
            lastMi--;
          }
        }
        E[i] = multiply(curPrefSum, inv(power(2LL, i + 1, mod)), mod);
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans = add(ans, E[i], mod);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) solve();
  return 0;
}
