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>
const double eps = 1e-8;
const int MAXN = 4e5 + 10;
const int mod = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n;
long long T, sum[MAXN];
int a[MAXN];
int p[MAXN], sz;
typedef struct node {
  int id;
  long long n, m;
} node;
node d[MAXN * 2];
bool cmp(node x, node y) {
  if (p[x.n] == p[y.n])
    return x.m < y.m;
  else
    return p[x.n] < p[y.n];
}
long long vip[MAXN], fvip[MAXN];
long long ksm(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y = y >> 1;
  }
  return ans;
}
long long ipv[MAXN], fipv[MAXN];
void init() {
  vip[0] = fvip[0] = 1;
  ipv[0] = fipv[0] = 1;
  for (int i = 1; i <= n; i++)
    vip[i] = vip[i - 1] * i % mod, fvip[i] = ksm(vip[i], mod - 2);
  for (int i = 1; i <= n; i++)
    ipv[i] = ipv[i - 1] * 2 % mod, fipv[i] = ksm(ipv[i], mod - 2);
}
long long ans1[MAXN], Vip;
long long C(long long x, long long y) {
  if (y < 0) return 0;
  assert(x >= y);
  return vip[x] * fvip[y] % mod * fvip[x - y] % mod;
}
int main() {
  n = read();
  T = read();
  sz = sqrt(n);
  Vip = ksm(2, mod - 2);
  init();
  for (int i = 1; i <= n; i++) a[i] = read(), sum[i] = sum[i - 1] + a[i];
  for (int i = 1; i <= n; i++) p[i] = (i - 1) / sz + 1;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    d[++cnt].id = i;
    d[cnt].n = i - 1;
    d[cnt].m = min((i - 1) * 1ll, T - sum[i - 1]);
    d[++cnt].id = -i;
    d[cnt].n = i;
    d[cnt].m = min(i * 1ll, T - sum[i]);
  }
  d[++cnt].id = n + 1;
  d[cnt].n = n;
  d[cnt].m = min(n * 1ll, T - sum[n]);
  sort(d + 1, d + cnt + 1, cmp);
  int L = 1;
  int R = 0;
  long long ans = 1;
  for (int i = 1; i <= cnt; i++) {
    if (d[i].m < 0) continue;
    while (L < d[i].n) {
      ans = (2ll * ans - C(L, R) + mod) % mod;
      L++;
    }
    while (L > d[i].n) {
      L--;
      ans = (ans + C(L, R)) % mod * Vip % mod;
    }
    while (R < d[i].m) {
      R++;
      ans = (ans + C(L, R)) % mod;
    }
    while (R > d[i].m) {
      ans = (ans - C(L, R) + mod) % mod;
      R--;
    }
    if (d[i].id < 0)
      ans1[-d[i].id] -= (ans * fipv[-d[i].id] % mod);
    else
      ans1[d[i].id] += (ans * fipv[d[i].id - 1] % mod);
  }
  for (int i = 1; i <= n + 1; i++) ans1[i] = (ans1[i] + mod) % mod;
  ans = 0;
  for (int i = 1; i <= n + 1; i++) ans += ans1[i] * (i - 1) % mod, ans %= mod;
  printf("%lld\n", ans);
}
