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;
template <typename T>
using V = vector<T>;
template <typename T, typename U>
using P = pair<T, U>;
template <typename T>
void cout_join(vector<T> &v, string d = " ") {
  for (int i = (0); i < (v.size()); ++i) {
    if (i > 0) cout << d;
    cout << v[i];
  }
  cout << endl;
}
template <typename T>
long long int power(T n, T p, T m) {
  if (p == 0) return 1LL;
  if (p == 1) return n;
  long long int k = power(n, p / 2, m);
  return ((k * k) % m * (p % 2 ? n : 1)) % m;
}
const long long int mod = 1e9 + 7;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  vector<long long int> sum(n + 1, 0);
  for (int i = (1); i <= (n); ++i) {
    sum[i] = sum[i - 1] + (s[i - 1] - '0');
  }
  for (int i = (0); i < (q); ++i) {
    long long int l, r;
    cin >> l >> r;
    long long int cnt1 = sum[r] - sum[l - 1], cnt0 = r - (l - 1) - cnt1;
    long long int enj = (power(2LL, cnt1, mod) - 1 + mod) % mod;
    enj += (enj * ((power(2LL, cnt0, mod) - 1 + mod) % mod)) % mod;
    enj %= mod;
    cout << enj << endl;
  }
  return 0;
}