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.
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 5, MOD = 1e9 + 7, eps = 1e-7, INF = 1e9;
const double PI = acos(-1);
int n, q, a, b;
string str;
int acum[maxN];
long long mulmod(long long a, long long b) {
  long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % MOD;
    b >>= 1, a = (a << 1) % MOD;
  }
  return ret;
}
long long fastPow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * x % MOD;
    n >>= 1, x = x * x % MOD;
  }
  return ret;
}
int main() {
  cin >> n >> q >> str;
  acum[0] = 0;
  for (int i = int(0); i < int(n); i++) {
    char c = str[i];
    if (c == '0') {
      acum[i + 1] = acum[i];
    } else {
      acum[i + 1] = acum[i] + 1;
    }
  }
  for (int i = int(0); i < int(q); i++) {
    cin >> a >> b;
    long long tot = b - a + 1, ceros = tot - (acum[b] - acum[a - 1]);
    long long ans = (fastPow(2, tot) + MOD - 1) % MOD;
    ans -= (fastPow(2, ceros) + MOD - 1) % MOD;
    ans += MOD;
    ans %= MOD;
    cout << ans << '\n';
  }
  return 0;
}
