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;
const long long int no = 3e6 + 5, modulo = 1e9 + 7, inf = 1e18, N = 3e3 + 1;
long long int ar[no], br[no];
long long int used[no];
long long int mul(long long int x, long long int y, long long int mod) {
  return ((x % mod) * (y % mod)) % mod;
}
long long int powwmod(long long int x, long long int y, long long int mod) {
  long long int res = 1;
  while (y) {
    if (y & 1) {
      y--;
      res = mul(res, x, mod);
      res %= mod;
    } else {
      y /= 2;
      x = mul(x, x, mod);
      x %= mod;
    }
  }
  return res % mod;
}
long long int calc(long long int c1, long long int c0) {
  long long int x = powwmod(2, c1, modulo);
  x -= 1;
  x = (x + mul(x, (powwmod(2, c0, modulo) - 1), modulo)) % modulo;
  return x;
}
void solve() {
  long long int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, x = 0, y = 0, z = 0,
                w = 0, k = 0;
  cin >> n;
  long long int q;
  cin >> q;
  string s;
  cin >> s;
  for (long long int i = 0; i < n; i++) {
    ar[i + 1] = ar[i] + s[i] - '0';
  }
  while (q--) {
    cin >> x >> y;
    z = calc(ar[y] - ar[x - 1], y - x + 1 - ar[y] + ar[x - 1]);
    cout << z << "\n";
  }
}
inline void runn() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  for (long long int i = 1; i < t + 1; i++) {
    solve();
  }
}