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 PYTHON3):

Solution:
 ######      ###      #######    #######    ##      #     #####        ###     ##### 
 #     #    #   #          #        #       # #     #    #     #      #   #     ###  
 #     #   #     #        #         #       #  #    #   #       #    #     #    ###  
 ######   #########      #          #       #   #   #   #           #########    #   
 ######   #########     #           #       #    #  #   #           #########    #   
 #     #  #       #    #            #       #     # #   #    ####   #       #    #   
 #     #  #       #   #             #       #      ##   #    #  #   #       #        
 ######   #       #  #######     #######    #       #    #####  #   #       #    #   
 
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key  # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
from bisect import *
from heapq import *
from math import log2
 
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x)) 
mod = int(1e9)+7
inf = float("inf")

n, q = gil()
a = gbs()

sp = [[0 for _ in range(n)] for _ in range(int(log2(n))+1)]

sp[0] = a
m = len(sp)

for i in range(1, m):
    d = 1<<(i-1)
    for j in range(n):
        sp[i][j] = sp[i-1][j] + (sp[i-1][j+d] if j+d < n else 0)


def getSum(l, r):
    ans = 0
    while l <= r:
        bt = int(log2(r-l+1))
        ans += sp[bt][l]
        delta = (1<<bt)
        l += delta


    return ans


for _ in range(q):
    l, r = gil()
    one = getSum(l-1, r-1)
    zero = r-l+1 - one
    ans = ((pow(2, one, mod) - 1 + mod)%mod)*pow(2, zero, mod)
    if ans >= mod:
        ans %= mod
    print(ans)