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.io.*;
import java.text.*;
import java.util.*;
import java.math.*;
 
public class template {
	public static void main(String[] args) throws Exception {
		new template().run();
	}
	public void run() throws Exception {
		FastScanner f = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = f.nextInt(); long t = f.nextLong();
		long[] arr = new long[n];
		long[] fact = new long[n+1], inv = new long[n+1];
		fact[0] = inv[0] = 1;
		int[] cnt = new int[n];
		for(int i = 0; i < n; i++) {
			arr[i] = f.nextLong();
			fact[i+1] = fact[i] * (i+1) % MOD;
			inv[i+1] = modInv(fact[i+1]);
			if(i != 0) arr[i] += arr[i-1];
			cnt[i] = (int) Math.min(t-arr[i], i+1);
		}
		long ans = 0;
		long psum = 1;
		int pcnt = 0;
		long cur = 500000004; 
		for(int i = 0; i < n; i++) {
			if(cnt[i] < 0) break;
			while(pcnt < cnt[i]) {
				pcnt++;
				psum = (psum + fact[i+1]*inv[pcnt]%MOD*inv[i+1-pcnt]%MOD)%MOD;
			}
			while(pcnt > cnt[i]) {
				psum = (psum - fact[i+1]*inv[pcnt]%MOD*inv[i+1-pcnt]%MOD+MOD)%MOD;
				pcnt--;
			}
			ans = (ans + cur * psum) % MOD;
			cur = cur * 500000004 % MOD;
			psum = psum * 2 % MOD;
			pcnt++;
			if(cnt[i] != i+1) psum = (psum + fact[i+1]*inv[pcnt]%MOD*inv[i+1-pcnt]%MOD) % MOD;
		}
		out.println(ans);
///
		out.flush(); 
	}
	long MOD = 1000000007, x, y;
	public long modInv(long a) {
		gcd(a, MOD);
		return (x%MOD+MOD)%MOD;
	}
	public void gcd(long a, long b) {
		if(a == 0) {
			x = 0; y = 1;
			return;
		}
		gcd(b%a,a);
		long nx = y - b/a*x;
		long ny = x;
		x = nx;
		y = ny;
		return;
	}
///
    static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }
}
