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 + 100;
const long long mod = 1e9 + 7;
long long m[maxn], pre[maxn];
char s[maxn];
int sum[maxn];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  m[0] = 1;
  long long base = 1;
  for (int i = 1; i <= n; i++) {
    base *= 2;
    base %= mod;
    m[i] = (m[i - 1] + base) % mod;
  }
  pre[0] = 1;
  for (int i = 1; i <= n; i++) pre[i] = (pre[i - 1] + m[i]) % mod;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1];
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int high = r - l - 1;
    int low = r - l - sum[r] + sum[l - 1];
    if (l == r) {
      if (s[l] == '1')
        printf("1\n");
      else
        printf("0\n");
      continue;
    }
    if (sum[r] - sum[l - 1] == 0)
      printf("0\n");
    else {
      int num = sum[r] - sum[l - 1];
      if (low <= 0)
        printf("%lld\n", (pre[high] + num) % mod);
      else
        printf("%lld\n", ((pre[high] - pre[low - 1] + mod) % mod + num) % mod);
    }
  }
  return 0;
}