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>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return ans % m;
}
const long double eps = 1e-10L;
long long one[300005], zero[300005];
vector<long long> v;
void pw() {
  long long val = 1;
  long long temp = 1;
  v.push_back(val);
  for (long long i = 1; i <= 300005; i++) {
    temp *= 2;
    temp %= (long long)(1000 * 1000 * 1000 + 7);
    val += temp;
    val %= (long long)(1000 * 1000 * 1000 + 7);
    v.push_back(val);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, q;
  cin >> n >> q;
  pw();
  string s;
  cin >> s;
  for (long long i = 1; i <= n; i++) {
    if (s[i - 1] == '1')
      one[i]++;
    else
      zero[i]++;
    one[i] += one[i - 1];
    zero[i] += zero[i - 1];
  }
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long o = one[b] - one[a - 1];
    long long z = zero[b] - zero[a - 1];
    long long ans = v[o + z - 1];
    if (z > 0) ans -= v[z - 1];
    ans += (long long)(1000 * 1000 * 1000 + 7);
    ans %= (long long)(1000 * 1000 * 1000 + 7);
    cout << ans << "\n";
  }
}