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 int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int n;
ll T, sum[maxn], t[maxn];
ll ni[maxn], jc[maxn];
ll qp(ll a, ll b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
ll C(int x, int y) {
  if (x < y) return 0;
  return jc[x] * ni[y] % mod * ni[x - y] % mod;
}
ll p2[maxn], rev2[maxn];
int main() {
  jc[0] = 1;
  for (int i = 1; i <= maxn - 1; i++) jc[i] = jc[i - 1] * i % mod;
  ni[maxn - 1] = qp(jc[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) ni[i] = ni[i + 1] * (i + 1) % mod;
  p2[0] = 1;
  for (int i = 1; i <= maxn - 1; i++) p2[i] = p2[i - 1] * 2 % mod;
  rev2[maxn - 1] = qp(p2[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) rev2[i] = rev2[i + 1] * 2 % mod;
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    sum[i] = sum[i - 1] + t[i];
  }
  t[n + 1] = sum[n + 1] = 1e18;
  ll s1 = 1, k1 = 0, s2 = 1, k2 = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    if (sum[i] > T) break;
    s1 = 2 * s1 + C(i - 1, k1 + 1);
    k1++;
    s2 = 2 * s2 + C(i - 1, k2 + 1);
    k2++;
    ll x = max(T - t[i + 1] + 1, sum[i]) - sum[i];
    ll y = min(1LL * i, T - sum[i]);
    while (k1 >= x) s1 -= C(i, k1--);
    while (k2 > y) s2 -= C(i, k2--);
    s1 %= mod;
    s2 %= mod;
    ll p = 0;
    if (x <= y) p += (s2 - s1) * rev2[i];
    if (x - 1 <= y && x - 1 >= 0) p += C(i, x - 1) * rev2[i + 1];
    p %= mod;
    (ans += p * i) %= mod;
  }
  cout << (ans % mod + mod) % mod << endl;
  return 0;
}
