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;
int dcmp(long double n, long double y) {
  return fabs(n - y) <= 1e-9 ? 0 : n < y ? -1 : 1;
}
const int MAX = 1e5 + 10;
const long long MOD = 1e9 + 7;
long long prefix[MAX][2], sum_of_pow[MAX], n, q;
string in;
long long POW_M(long long a, long long p, long long m = MOD) {
  if (p == 0) return 1;
  if (p == 1) return a % m;
  long long x = POW_M(a, p / 2, m);
  if (p % 2 == 0) return ((x % m) * x) % m;
  return (((x % m) * x % m) * a) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> q;
  sum_of_pow[0] = 1;
  cin >> in;
  for (long long i = 1; i < MAX; i++) {
    sum_of_pow[i] = (2LL * sum_of_pow[i - 1]) % MOD;
  }
  for (int i = 0; i < (int)(MAX - 2); ++i) {
    sum_of_pow[i + 1] = (sum_of_pow[i + 1] % MOD + sum_of_pow[i] % MOD) % MOD;
  }
  for (int i = 0; i < ((int)((in).size())); ++i) {
    prefix[i + 1][0] += (in[i] == '0') + prefix[i][0];
    prefix[i + 1][1] += (in[i] == '1') + prefix[i][1];
  }
  for (int i = 0; i < (int)(q); ++i) {
    int l, r;
    cin >> l >> r;
    long long z = prefix[r][0] - prefix[l - 1][0];
    long long o = prefix[r][1] - prefix[l - 1][1];
    long long ans = 0;
    if (o > 0) {
      ans += sum_of_pow[o - 1];
    }
    if (z > 0) {
      ans +=
          (o > 0 ? (sum_of_pow[z - 1] * (sum_of_pow[o - 1] % MOD)) % MOD : 0LL);
      ans %= MOD;
    }
    cout << ans << '\n';
  }
}