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.
import java.util.Scanner;

// 2500
public class CrosswordExpert1194F {

    static final long MOD = 1_000_000_007;

    static long n;
    static long T;
    static long[] t; // [0] = 0, [1:] = input, [n + 1] = T


    public static void main(String[] args) throws Exception {
        try (Scanner scanner = new Scanner(System.in)) {
            n = scanner.nextLong();
            T = scanner.nextLong();
            t = new long[(int) n + 2];
            t[(int) n + 1] = T;
            for (int i = 1; i <= n; i++) t[i] = scanner.nextLong();
            System.out.println(mean());
        }
    }

    static long mean() {
        long res = 0;
        long sum = 0;
        // choose(x, y) = x! / (y! * (x - y)!)
        // choose(x, y - 1) = choose(x, y) * y / (x - y + 1)
        // choose(x + 1, y) = choose(x, y) * (x + 1) / (x - y + 1)
        // choose(x + 1, y + 1) = choose(x, y) * (x + 1) / (y + 1)
        long choose = 1; // choose(x, y)
        for (int x = 0, y = 0; x <= n && y >= 0;) {
            if (sum + time(x + 1, y + 1) <= T) {
                sum += time(x + 1, 0);
                choose = mul(choose, x + 1, pow(y + 1, -1));
                x++;
                y++;
            } else if (sum + time(x + 1, y) <= T) {
                res = add(res, mul(x, pow(2, -x - 1), choose));
                sum += time(x + 1, 0);
                choose = mul(choose, x + 1, pow(x - y + 1, -1));
                x++;
            } else {
                res = add(res, mul(x, pow(2, -x), choose));
                choose = mul(choose, y, pow(x - y + 1, -1));
                y--;
            }
        }
        return res;
    }

    static long time(int x, int y) {
        return t[x] + y;
    }

    static long mod(long a) {
        return a < 0 ? mod(a + MOD)
                : a < MOD ? a
                : a % MOD;
    }

    static long add(long a, long b) {
        return mod(a + b);
    }

    static long mul(long a, long b) {
        return mod(mod(a) * mod(b));
    }

    static long mul(long... values) {
        long res = 1;
        for (long v : values) res = mul(res, v);
        return res;
    }

    static long pow(long a, long b) {
        if (b < 0) return pow(a, MOD + b - 1);
        long n = b;
        long x = a;
        long res = 1;
        while (n > 0) {
            if ((n & 1) != 0) res = mul(res, x);
            x = mul(x, x);
            n >>= 1;
        }
        return res;
    }
}
