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 N = 200005;
const int mod = 1e9 + 7;
ll fexp(ll b, ll e = mod - 2) {
  ll ans = 1;
  while (e) {
    if (e & 1) ans = ans * b % mod;
    b = b * b % mod;
    e /= 2;
  }
  return ans;
}
ll fat[N], ifat[N];
ll C(int n, int k) {
  if (k < 0 or k > n) return 0;
  return fat[n] * ifat[k] % mod * ifat[n - k] % mod;
}
int n;
ll T, t[N], p[N];
ll val = 1;
ll L = 0, R = 0;
ll vn = 0;
void addN() {
  val = (2 * val + C(vn, L - 1) - C(vn, R) + mod) % mod;
  vn++;
}
void fix(ll nR) {
  while (R < nR) {
    R++;
    val += C(vn, R);
    if (val >= mod) val -= mod;
  }
  while (R > nR) {
    val -= C(vn, R);
    if (val < 0) val += mod;
    R--;
  }
}
int main() {
  fat[0] = ifat[0] = 1;
  for (int i = 1; i < N; i++) {
    fat[i] = i * fat[i - 1] % mod;
    ifat[i] = fexp(fat[i]);
  }
  scanf("%d %lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%lld", t + i);
  partial_sum(t, t + n + 1, p);
  t[n + 1] = 2 * T;
  ll ans = 0;
  for (int i = 1; i <= n; i++) {
    addN();
    ll mxAdd = T - p[i];
    if (mxAdd < 0) break;
    if (mxAdd > i) mxAdd = i;
    fix(mxAdd);
    ans = (ans + val * fexp(fexp(2, i))) % mod;
  }
  printf("%lld\n", ans);
}
