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.
//package ContestEd68;

import java.io.*;
import java.util.StringTokenizer;

public class mainF {
    public static PrintWriter out = new PrintWriter(System.out);
    public static FastScanner enter = new FastScanner(System.in);
    public static long fact[];
    public static long mod = (long) (1e9) + 7;
    public static long dub = bin_pow(2, mod - 2, mod);

    public static void main(String[] args) throws IOException {
        int n = enter.nextInt();
        long T = enter.nextLong();
        fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i;
            fact[i] %= mod;
        }
        long t[] = new long[n + 1];
        for (int i = 1; i <= n; i++) {
            t[i] = enter.nextLong();
        }
        for (int i = 1; i < n + 1; i++) {
            t[i] += t[i - 1];
        }
        for (int i = 1; i < n + 1; i++) {
            t[i] = T - t[i];
        }
        int c = 1;
        while (c < t.length && t[c] >= c) c++;
        if (c == t.length) {
            System.out.println(n);
            return;
        }
        long[] mass = new long[n + 1];
        for (int i = 0; i <= t[c]; i++) {
            mass[c] += cnk(c, i);
            mass[c] %= mod;
        }
        for (int i = c + 1; i < n + 1; i++) {
            if (t[i] < 0) break;
            long tmp = mass[i - 1] * 2 + cnk(i-1 , (int) t[i - 1]+1);
            tmp %= mod;
            int up = (int) t[i-1]+1;
            while (up != t[i]) {
                tmp -= cnk(i, up);
                up--;
                tmp %= mod;
            }
            mass[i] = tmp;
        }
        long ans = 0;
        for (int i = 1; i < n + 1; i++) {
            if (t[i] < 0) break;
            if (t[i] >= i) {
                ans++;
                continue;
            }
            ans += mass[i] * bin_pow(dub, i, mod);
            ans %= mod;
        }
        ans = (ans % mod + mod) % mod;
        System.out.println(ans);
    }

    public static long cnk(int n, int k) {
        long ans = fact[n];
        long tmp = fact[n - k] * fact[k];
        tmp %= mod;
        tmp = bin_pow(tmp, mod - 2, mod);
        ans *= tmp;
        ans %= mod;
        return ans;
    }

    public static long bin_pow(long a, long b, long mod) {//a^b %mod
        long ans = 1;
        while (b != 0) {
            if ((b & 1) == 1) ans *= a;
            a *= a;
            ans %= mod;
            a %= mod;
            b >>= 1;
        }
        return ans;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer stok;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() throws IOException {
            while (stok == null || !stok.hasMoreTokens()) {
                String s = br.readLine();
                if (s == null) {
                    return null;
                }
                stok = new StringTokenizer(s);
            }
            return stok.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        char nextChar() throws IOException {
            return (char) (br.read());
        }

        String nextLine() throws IOException {
            return br.readLine();
        }
    }
}


