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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
using namespace std;
long long fastexpom(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (((res) % m) * ((a) % m)) % m;
    a = (((a) % m) * ((a) % m)) % m;
    b /= 2;
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  vector<long long> v(n + 1);
  vector<long long> cnt0(n + 1, 0);
  vector<long long> cnt1(n + 1, 0);
  for (long long i = 0; i < n; i++) {
    v[i + 1] = s[i] - '0';
    if (v[i + 1]) {
      cnt1[i + 1] = cnt1[i] + 1;
      cnt0[i + 1] = cnt0[i];
    } else {
      cnt0[i + 1] = cnt0[i] + 1;
      cnt1[i + 1] = cnt1[i];
    }
  }
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long one = cnt1[r] - cnt1[l - 1];
    long long zero = cnt0[r] - cnt0[l - 1];
    if (one == 0) {
      cout << "0"
           << "\n";
      continue;
    }
    long long half =
        (fastexpom(2, one, 1000000007) - 1 + 1000000007) % 1000000007;
    long long start = half;
    long long ans = half;
    if (zero != 0) {
      half = (fastexpom(2, zero, 1000000007) - 1 + 1000000007) % 1000000007;
      half = half * start;
      half = half % 1000000007;
      ans = (ans + half) % 1000000007;
    }
    cout << ans << "\n";
  }
  return 0;
}