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 MOD = 1e9 + 7;
const int MN = 2e5 + 100;
int N;
ll T;
ll f[2][MN];
ll inv[MN];
ll p2[2][MN];
void pcp(void) {
  f[0][0] = f[1][0] = 1LL;
  inv[1] = 1;
  for (int i = 2; i < MN; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  for (int i = 1; i < MN; i++)
    f[0][i] = f[0][i - 1] * i % MOD, f[1][i] = f[1][i - 1] * inv[i] % MOD;
  p2[0][0] = p2[1][0] = 1LL;
  for (int i = 1; i < MN; i++)
    p2[0][i] = p2[0][i - 1] * 2 % MOD, p2[1][i] = p2[1][i - 1] * inv[2] % MOD;
}
ll choose(int n, int k) {
  if (k > n)
    return 0;
  else
    return f[0][n] * f[1][k] % MOD * f[1][n - k] % MOD;
}
ll ans;
ll sm, t;
ll pn = 0, pk = 0, pv = 1;
void inc(void) {
  pv = (pv * 2 + choose(pn, pk + 1) + MOD) % MOD;
  pn++, pk++;
}
ll pick(ll n, ll k) {
  for (; pn < n; inc())
    ;
  for (; pk > k; pk--) pv = (pv - choose(n, pk)) % MOD;
  return pv;
}
int main(void) {
  pcp();
  scanf("%d%lld", &N, &T);
  for (int i = 1; i <= N; i++) {
    scanf("%lld", &t);
    sm += t;
    if (sm + i <= T)
      ans++;
    else if (sm > T)
      break;
    else
      ans = (ans + (pick(i, T - sm) + MOD) % MOD * p2[1][i] % MOD) % MOD;
  }
  printf("%lld\n", ans % MOD);
  return 0;
}
