Problem: 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.

Solution (in JAVA):

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


public class Main {
    static final int mod = (int) (1e9 + 7);

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt();
        int q = sc.nextInt();
        char[] x = sc.next().toCharArray();
        FenwickTree ft = new FenwickTree(n);
        for (int i = 1; i <= n; i++)
            ft.point_update(i, x[i - 1] - '0');
        while (q-- > 0) {
            int i = sc.nextInt();
            int j = sc.nextInt();
            int ones = ft.rsq(i, j);
            int zeros = (j - i + 1) - ones;
            int ans = gs(1, 2, ones) % mod;
            ans += gs(ans, 2, zeros);
            out.println(ans % mod);
        }
        out.flush();
        out.close();
    }

    static int gs(int a, int r, int n) {
        int ans = (int) (1 - modPow(r, n, mod)) % mod;
        ans /= (1 - r);
        ans = (int) ((1l * a * ans) % mod);
        return ans % mod;
    }

    static int modPow(int a, int e, int mod)    // O(log e)
    {
        a %= mod;
        long res = 1;
        while (e > 0) {
            if ((e & 1) == 1)
                res = (1l*res * a) % mod;
            a = (int) ((1l*a * a) % mod);
            e >>= 1;
        }
        return (int) (res%mod);
    }

    static class FenwickTree { // one-based DS

        int n;
        int[] ft;

        FenwickTree(int size) {
            n = size;
            ft = new int[n + 1];
        }

        int rsq(int b) //O(log n)
        {
            int sum = 0;
            while (b > 0) {
                sum += ft[b];
                b -= b & -b;
            }        //min?
            return sum;
        }

        int rsq(int a, int b) {
            return rsq(b) - rsq(a - 1);
        }

        void point_update(int k, int val)    //O(log n), update = increment
        {
            while (k <= n) {
                ft[k] += val;
                k += k & -k;
            }        //min?
        }

        int point_query(int idx)    // c * O(log n), c < 1
        {
            int sum = ft[idx];
            if (idx > 0) {
                int z = idx ^ (idx & -idx);
                --idx;
                while (idx != z) {
                    sum -= ft[idx];
                    idx ^= idx & -idx;
                }
            }
            return sum;
        }

        void scale(int c) {
            for (int i = 1; i <= n; ++i) ft[i] *= c;
        }

        int findIndex(int cumFreq) {
            int msk = n;
            while ((msk & (msk - 1)) != 0)
                msk ^= msk & -msk;            //msk will contain the MSB of n

            int idx = 0;
            while (msk != 0) {
                int tIdx = idx + msk;
                if (tIdx <= n && cumFreq >= ft[tIdx]) {
                    idx = tIdx;
                    cumFreq -= ft[tIdx];
                }
                msk >>= 1;
            }
            if (cumFreq != 0)
                return -1;
            return idx;
        }
    }

    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        Scanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

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

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

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

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


}