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>
long long n, t, i, j, arr[1000004], a, b, store, arr1[1000004], ans,
    final[1000004], ans1;
long long aa, mod = 1000000007, zero[1000004], one[1000004], zer, on,
              final1[1000004];
char name[1000004];
using namespace std;
int main() {
  scanf("%lld %lld", &n, &t);
  scanf("%s", name);
  store = 0;
  ans = 0;
  for (i = 1; i <= 1000000; i++) {
    ans = (ans + store + 1) % mod;
    final[i] = ans;
    store = ans % mod;
  }
  aa = 2;
  final[0] = 0;
  final1[0] = 1;
  for (i = 1; i <= 1000000; i++) {
    if (i == 1) {
      final1[i] = 2;
    } else {
      final1[i] = ((final1[i - 1] % mod) * 2) % mod;
    }
  }
  for (i = 0; i < n; i++) {
    arr[i] = (int)name[i] - '0';
  }
  for (i = 0; i < n; i++) {
    if (i == 0) {
      if (arr[0] == 0) {
        zero[0] = 1;
        one[0] = 0;
      } else {
        zero[0] = 0;
        one[0] = 1;
      }
    } else {
      if (arr[i] == 1) {
        zero[i] = zero[i - 1];
        one[i] = one[i - 1] + 1;
      } else {
        zero[i] = zero[i - 1] + 1;
        one[i] = one[i - 1];
      }
    }
  }
  while (t--) {
    ans = 0;
    j = 0;
    scanf("%lld %lld", &a, &b);
    if (a == 1) {
      zer = zero[b - 1];
      on = one[b - 1];
    } else {
      zer = zero[b - 1] - zero[a - 2];
      on = one[b - 1] - one[a - 2];
    }
    ans = final[on];
    ans1 = final1[zer];
    ans = ((ans1 % mod) * (ans % mod)) % mod;
    printf("%lld\n", ans);
  }
}