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 lli = long long;
const int MOD = 1e9 + 7;
lli power(lli a, lli p) {
  lli ret = 1;
  while (p) {
    if (p & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    p >>= 1;
  }
  return ret;
}
lli inv(lli a) { return power(a, MOD - 2); }
int n;
lli arr[200002];
lli T, fact[200002];
int check(lli a, int idx) {
  int lo = -1, up = idx + 1;
  while (up - lo > 1) {
    int md = (lo + up) >> 1;
    if (a + md > T)
      up = md;
    else
      lo = md;
  }
  return lo;
}
int main() {
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%lld", arr + i), arr[i] += arr[i - 1];
  arr[n + 1] = T + 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % MOD;
  lli ans = 0;
  for (int i = 1; i <= n; i++) {
    int k = check(arr[i], i);
    for (int j = k; j >= 0 && arr[i + 1] + j > T; j--) {
      lli prob = fact[i] * inv(fact[j]) % MOD * inv(fact[i - j]) % MOD *
                 inv(power(2, i)) % MOD;
      ans = (ans + prob * inv(2) % MOD * i % MOD) % MOD;
    }
    for (int j = k; j >= 0 && arr[i + 1] + j + 1 > T; j--) {
      lli prob = fact[i] * inv(fact[j]) % MOD * inv(fact[i - j]) % MOD *
                 inv(power(2, i)) % MOD;
      ans = (ans + prob * inv(2) % MOD * i % MOD) % MOD;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
