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;
void no() { cout << "NO\n"; }
void yes() { cout << "YES\n"; }
long long int pow_(long long int a, long long int n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n % 2 == 0)
    return ((pow_(a, n / 2) % 1000000007) * (pow_(a, n / 2) % 1000000007)) %
           1000000007;
  else
    return ((((pow_(a, n / 2) % 1000000007) * (pow_(a, n / 2) % 1000000007)) %
             1000000007) *
            (a % 1000000007)) %
           1000000007;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, q;
  string s;
  cin >> n >> q >> s;
  long long int o = 0;
  long long int z = 0;
  vector<pair<long long int, long long int> > v(n + 1);
  v[0] = make_pair(0, 0);
  for (long long int i = 0; i < n; i += 1) {
    if (s[i] == '1') o++;
    if (s[i] == '0') z++;
    v[i + 1] = make_pair(o, z);
  }
  for (long long int i = 0; i < q; i += 1) {
    long long int l, r;
    cin >> l >> r;
    long long int len = r - l + 1, ones = v[r].first - v[l - 1].first;
    long long int a = pow_(2, len - ones) % 1000000007,
                  b = ((pow_(2, ones) % 1000000007) - (1 % 1000000007)) %
                      1000000007;
    if (b < 0) b += 1000000007;
    cout << ((a % 1000000007) * (b % 1000000007)) % 1000000007 << "\n";
  }
  return 0;
}