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;
namespace fast_IO {
inline int read_int() {
  register int ret = 0, f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = (ret << 1) + (ret << 3) + int(c - 48);
    c = getchar();
  }
  return ret * f;
}
}  // namespace fast_IO
using namespace fast_IO;
int N, Q;
namespace BIT {
int c[200005];
inline int lowbit(int x) { return x & (-x); }
inline void add(int pos, int w) {
  while (pos <= N) c[pos] += w, pos += lowbit(pos);
}
inline int query(int pos) {
  int res = 0;
  while (pos) res += c[pos], pos -= lowbit(pos);
  return res;
}
}  // namespace BIT
long long w[200005];
char s[200005];
inline void init() {
  N = read_int(), Q = read_int();
  scanf("%s", s + 1);
  for (register int i = 1; i <= N; i++) BIT::add(i, s[i] - '0');
  w[1] = 1;
  for (register int i = 2; i <= N + 1; i++)
    w[i] = w[i - 1] << 1, w[i] %= 1000000007;
  for (register int i = 1; i <= N + 1; i++)
    w[i] += w[i - 1], w[i] %= 1000000007;
  for (register int i = 1; i <= N + 1; i++)
    w[i] += w[i - 1], w[i] %= 1000000007;
}
inline void calc() {
  register int l, r;
  register int zero, one;
  while (Q--) {
    l = read_int(), r = read_int();
    one = BIT::query(r) - BIT::query(l - 1);
    zero = r - l + 1 - one;
    int len = r - l + 1;
    long long ans = w[len - 1] - w[len - one - 1] + one;
    ans = (ans % 1000000007 + 1000000007) % 1000000007;
    printf("%lld\n", ans);
  }
}
int main() {
  init();
  calc();
  return 0;
}