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 int maxn = 1e5 + 5;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e = exp(1);
const double PI = acos(-1);
const double ERR = 1e-10;
long long pow_(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % Mod;
    b >>= 1;
    a = (a * a) % Mod;
  }
  return ans;
}
long long inv(long long x) { return pow_(x, Mod - 2); }
long long sum[maxn][2];
char str[maxn];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", str + 1);
  for (int i = 1; i <= n; i++) {
    sum[i][0] = sum[i - 1][0];
    sum[i][1] = sum[i - 1][1];
    if (str[i] == '0')
      sum[i][0] = (sum[i][0] + 1) % Mod;
    else
      sum[i][1] = (sum[i][1] + 1) % Mod;
    ;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long sum0 = (sum[r][0] - sum[l - 1][0] + Mod) % Mod;
    long long sum1 = (sum[r][1] - sum[l - 1][1] + Mod) % Mod;
    long long ans = 0;
    ans = (pow_(2, sum1) - 1 + Mod) % Mod;
    long long st = (pow_(2, sum1) - 1 + Mod) % Mod;
    long long tmp = (pow_(2, sum0) - 1 + Mod) % Mod;
    tmp = (tmp * st) % Mod;
    ans = (ans + tmp) % Mod;
    printf("%lld\n", ans);
  }
  return 0;
}