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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
using ll = long long;
const ll mod = 1e9 + 7;
const int maxn = 1e6 + 5;
ll pw(ll x, ll y) {
  ll res = 1;
  while (y) {
    if (y % 2) res = res * x % mod;
    x = x * x % mod;
    y = y / 2;
  }
  return ((res % mod) + mod) % mod;
}
ll inv1(ll x) {
  assert(x > 0);
  return pw(x, mod - 2);
}
ll fac[maxn], ifac[maxn], inv[maxn];
ll nck(ll n, ll k) {
  if (n < k) return 0;
  return fac[n] * ifac[k] % mod * ifac[n - k] % mod;
}
void gen_inv() {
  fac[0] = inv[0] = ifac[0] = 1;
  fac[1] = inv[1] = ifac[1] = 1;
  for (int i = 2; i < maxn; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = (mod - (mod / i) * inv[mod % i] % mod) % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
  assert(nck(4, 2) == 6);
}
ll n, t;
ll a[maxn];
ll acc[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  gen_inv();
  cin >> n >> t;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    acc[i] = a[i] + acc[i - 1];
  }
  ll res = 0;
  ll curi = 0;
  ll curm = 0;
  ll choose = 1;
  {
    ll a = nck(5, 0) + nck(5, 1);
    ll b = 2 * a - nck(5, 0) - nck(5, 1) + nck(6, 0);
    assert(b == (nck(6, 0) + nck(6, 1)));
  }
  auto fix_choose = [&](ll i, ll m) {
    while (curi < i) {
      choose = choose * 2 % mod;
      choose -= nck(curi, 0);
      choose %= mod;
      choose -= nck(curi, curm);
      choose %= mod;
      choose += nck(curi + 1, 0);
      choose %= mod;
      curi++;
    }
    while (curm < m) {
      choose += nck(curi, ++curm);
      choose %= mod;
    }
    while (curm > m) {
      choose -= nck(curi, curm--);
      choose %= mod;
    }
    assert(curi == i && curm == m);
  };
  auto nck_sum = [&](ll i, ll m) {
    ll res = 0;
    for (ll k = 0; k <= m; k++) {
      res += nck(i, k);
      res %= mod;
    }
    return res;
  };
  for (int i = 1; i <= n; i++) {
    t -= a[i];
    if (t < 0) break;
    ll cur = 0;
    ll m = min((ll)i, t);
    fix_choose(i, m);
    cur += choose;
    cur %= mod;
    cur = cur * inv1(pw(2, i)) % mod;
    res += cur;
    res %= mod;
  }
  res %= mod;
  res += mod;
  res %= mod;
  cout << res << endl;
  {
    ll a = nck(5, 0) + nck(5, 1) + nck(5, 2);
    ll b = 2 * a - nck(5, 0) - nck(5, 2) + nck(6, 0);
    ll c = nck(6, 0) + nck(6, 1) + nck(6, 2);
    assert(b == c);
  }
  return 0;
}
