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.
import configparser
import math
import sys


input = sys.stdin.readline


def mod_exp(base, exp, mod):
    # cur_iter = base^(2^0)) = base^1 = base
    cur_iter = base
    cur_iter %= mod

    result = 1

    while exp > 0:
        if (1 & exp) > 0:
            result = result * cur_iter
            result %= mod

        cur_iter *= cur_iter
        cur_iter %= mod

        exp = exp >> 1

    return result % mod

def mod_mul(a, b, mod):
    return ((a%mod) * (b%mod))%mod

def mod_add(a, b, mod):
    return ((a%mod) + (b%mod))%mod

def main():

    mod = int(1e9) + 7

    n, q = [int(x) for x in input().split(' ')]
    s = input().strip()

    cnt0 = [0 for i in range(n)]
    cnt1 = [0 for i in range(n)]

    cnt0[0] = int((s[0]=='0'))
    cnt1[0] = int((s[0]=='1'))

    for i in range(1, n):
        cnt0[i] = (s[i]=='0') + cnt0[i-1]
        cnt1[i] = (s[i]=='1') + cnt1[i-1]

    for i in range(q):
        l, r = [int(x) for x in input().split(' ')]

        l -= 1
        r -= 1

        x = cnt0[r]
        if l != 0:
            x -= cnt0[l-1]

        y = cnt1[r]
        if l != 0:
            y -= cnt1[l-1]

        ans = mod_exp(2, y, mod) - 1

        if x > 0:
            ans += mod_mul(ans, mod_exp(2, x, mod)-1, mod)
            ans %= mod

        print(ans)


if __name__ == '__main__':
    main()
