JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.
//https://www.hackerearth.com/submission/5284067/

import java.io.*;
import java.lang.reflect.Array;
import java.util.*;

public class Main {

    public static int n, m;
    public static ArrayList<Integer>[] adj, idx;
    public static ArrayList<Long> num[];
    public static int chainNo, ptr;
    public static int[] chainHead, chainSize, chainIdx, posInBase, subtree;
    public static long[] baseArray, segtree;
    public static int[] depth, parent;
    public static int[][] P;
    public static int rev[];
    public static int LOGMAXN;
    public static int MOD=(int)(1e9+7);
    public static void main(String args[]) {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }, "1", 1 << 26).start();
    }

    @SuppressWarnings("unchecked")
    public static void solve() {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n=in.nextInt();
        int q=in.nextInt();
        int a[]=new int[n];
        String st=in.readString();
        for(int i=0;i<n;i++) {
            a[i]=st.charAt(i)-'0';
        }
        int pre[]=new int[n+1];
        for(int i=1;i<=n;i++) {
            pre[i]=pre[i-1]+a[i-1];
        }
        long pow[]=new long[n+1];
        pow[0]=1;
        for(int i=1;i<=n;i++){
            pow[i]=pow[i-1]*2;
            pow[i]%=MOD;
        }


        for(int i=0;i<q;i++){
            int l=in.nextInt();
            int r=in.nextInt();
            int length= r-l+1;
            int cnt= pre[r]-pre[l-1];
            if(cnt==0){
                out.println("0");
                continue;
            }
            long sum = (pow[cnt]-1+MOD)%MOD;
            int rem= length - cnt;
            if(rem>0){
                sum= sum+((sum*(pow[rem]-1+MOD)%MOD)%MOD);
                sum%=MOD;
            }
            out.println(sum);
        }


        out.close();
    }

    public static int KMPSearch(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();

        // create lps[] that will hold the longest
        // prefix suffix values for pattern
        int lps[] = new int[M];
        int j = 0; // index for pat[]

        // Preprocess the pattern (calculate lps[]
        // array)
        computeLPSArray(pat, M, lps);

        int i = 0; // index for txt[]
        while (i < N) {
            if (pat.charAt(j) == txt.charAt(i)) {
                j++;
                i++;
            }
            if (j == M) {
                return (i - j);
                //j = lps[j - 1];
            }

            // mismatch after j matches
            else if (i < N && pat.charAt(j) != txt.charAt(i)) {
                // Do not match lps[0..lps[j-1]] characters,
                // they will match anyway
                if (j != 0)
                    j = lps[j - 1];
                else
                    i = i + 1;
            }
        }
     return  -1;
    }

    public static void computeLPSArray(String pat, int M, int lps[])
    {
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0

        // the loop calculates lps[i] for i = 1 to M-1
        while (i < M) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            }
            else // (pat[i] != pat[len])
            {
                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar
                // to search step.
                if (len != 0) {
                    len = lps[len - 1];

                    // Also, note that we do not increment
                    // i here
                }
                else // if (len == 0)
                {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }







    static class InputReader {

        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int snumChars;
        private SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int snext() {
            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = snext();
            while (isSpaceChar(c))
                c = snext();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = snext();
            }

            int res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = snext();
            } while (!isSpaceChar(c));

            return res * sgn;
        }

        public long nextLong() {
            int c = snext();
            while (isSpaceChar(c))
                c = snext();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = snext();
            }

            long res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = snext();
            } while (!isSpaceChar(c));

            return res * sgn;
        }

        public String readString() {
            int c = snext();
            while (isSpaceChar(c))
                c = snext();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = snext();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);
        }
    }
}