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;
long long dx[] = {-1, 1, 0, 0};
long long dy[] = {0, 0, -1, 1};
long long sub(long long a, long long b, long long p = 1000000007) {
  return ((a % p) - (b % p) + p) % p;
}
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  while (k > 0) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
    k = k >> 1;
  }
  return r;
}
inline long long powr(long long a, long long b) {
  long long x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long fdiv(long long a, long long b, long long p = 1000000007) {
  long long yinv = inv(b);
  long long ans = (a * yinv) % p;
  return ans;
}
long long _pow(long long p, long long q) {
  long long res = 1;
  while (q) {
    if (q & 1) res = (res * p) % 1000000007;
    p = (p * p) % 1000000007;
    q = q >> 1;
  }
  return (res + 1000000007) % 1000000007;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  vector<long long> ones(n + 1);
  for (long long i = 0; i < n; ++i) {
    if (s[i] == '1')
      ones[i + 1] = ones[i] + 1;
    else
      ones[i + 1] = ones[i];
  }
  for (long long i = 0; i < q; ++i) {
    long long l, r;
    cin >> l >> r;
    long long one = ones[r] - ones[l - 1];
    long long z = r - l + 1 - one;
    long long p = _pow(2ll, one) - 1;
    long long k = mult(z, z + 1) / 2;
    long long ans = add(p, mult((_pow(2ll, z) - 1), p));
    cout << ans << "\n";
  }
}