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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char str[1008611];
long long s1[1008611], s2[1008611];
long long pow_(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    b >>= 1;
    a = (a * a) % 1000000007;
  }
  return ans;
}
long long inv(long long x) { return pow_(x, 1000000007 - 2); }
int main() {
  int n, q;
  cin >> n >> q;
  cin >> str;
  if (str[0] == '0') {
    s1[0] = 1;
    s2[0] = 0;
  } else {
    s1[0] = 0;
    s2[0] = 1;
  }
  for (int i = 1; i < n; i++) {
    s1[i] = s1[i - 1];
    s2[i] = s2[i - 1];
    if (str[i] == '0') {
      s1[i] = (s1[i] + 1) % 1000000007;
    } else {
      s2[i] = (s2[i] + 1) % 1000000007;
    }
  }
  long long n1, n2;
  while (q--) {
    int l, r;
    cin >> l >> r;
    n1 = (s1[r - 1] - s1[l - 1]) % 1000000007;
    n2 = (s2[r - 1] - s2[l - 1]) % 1000000007;
    if (str[l - 1] == '0') {
      n1 = (n1 + 1) % 1000000007;
    } else {
      n2 = (n2 + 1) % 1000000007;
    }
    long long a, b;
    a = pow_(2, n2) - 1;
    a = a % 1000000007;
    b = ((pow_(2, n1) - 1) * a);
    b = (b + 1000000007) % 1000000007;
    long long sum = (a + b) % 1000000007;
    cout << sum << endl;
  }
  return 0;
}