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 = 200010;
const long long mo = 1e9 + 7;
int n, m, k;
long long a[maxn], sum[maxn];
long long c[maxn];
long long ans, ct, cnt, tmp, flag;
char s[maxn];
long long power(long long a, long long n) {
  long long ans = 1;
  a = a % mo;
  while (n) {
    if (n & 1) ans = (ans * a) % mo;
    n >>= 1;
    a = (a * a) % mo;
  }
  return ans;
}
int main() {
  int T, cas = 1;
  while (scanf("%d%d", &n, &m) != EOF) {
    ans = 0;
    flag = 1;
    memset(c, 0, sizeof(c));
    scanf("%s", s + 1);
    for (int i = 1; i <= n; i++) {
      if (s[i] == '0')
        c[i] = c[i - 1];
      else
        c[i] = c[i - 1] + 1;
    }
    while (m--) {
      long long x, y;
      scanf("%lld%lld", &x, &y);
      if (c[y] - c[x - 1] == 0) {
        puts("0");
        continue;
      }
      long long k = c[y] - c[x - 1];
      long long tp = (power(2, k) - 1 + mo) % mo;
      long long kk = (y - x + 1 - k);
      long long tmp = (tp * (power(2, kk) - 1 + mo) % mo) % mo;
      ans = (tmp + tp) % mo;
      printf("%lld\n", ans);
    }
  }
  return 0;
}